What name is given to "an object or group of objects that we wish to consider for a physics problem?​

Answer:

Gravity only becomes noticeable when there is a really massive object like a moon, planet or star. We are pulled down towards the ground because of gravity. The gravitational force pulls in the direction towards the centre of any object.

Answer:

B: 1530 N

Explanation:

We know ghat formula for force of gravity is;

F_grav = G•m1•m2/d²

We are told that two clay masses produce a gravitational force of 340N.

Thus;

G•m1•m2/d² = 340

Now, if the distance is divided by 3 and the mass of one is divided by 2, we have;

F_g = (G × m1/2 × m2)/(d/3)²

Thus gives;

F_g = ½(G•m1•m2)/((1/9)d²)

Simplifying this gives;

F_g = (9/2)G•m1•m2/d²

From earlier, we saw that;

G•m1•m2/d² = 340

Thus;

F_g = (9/2) × 340

F_g = 1530 N

Answer:

x = 1382.9 km

Explanation:

The speed of the wave is constant, so we can use the uniform motion relationships

p wave

          [tex]v_p[/tex] = x / t₁

          t₁ = x /v_p

S wave

          v_s = x / t₂

           t₂ = x / v_s

indicate that the time difference between the two waves is

          t₂ - t₁ = 3.25 min (60 s / 1 min)

          t₂ -t₁ = 195 s

let's substitute

         [tex]\frac{x}{v_s} - \frac{x}{v_p}[/tex] = 195

         x ([tex]\frac{1}{v_s} - \frac{1}{v_p}[/tex] = 195

let's calculate

         x [tex]( \frac{1}{4.2} - \frac{1}{10.3} )[/tex] = 195

         x (0.1410) = 195

         x = 195 /0.141

         x = 1382.9 km

Iron is magnetic, so any metal with iron in it will be attracted to a magnet. Steel contains iron, so a steel paperclip will be attracted to a magnet too. Most other metals, for example, aluminum, copper, and gold are NOT magnetic. Two metals that aren't magnetic are gold and silver.

Answer:

Kinetic energy =1/2mv^2

Where m is the mass of the object. V is the velocity

K.E=0.5(8*5^2)

K.E=100J

Explanation:

Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

1)

       [tex]F_{g} = F_{c} (1)[/tex]

       [tex]F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)[/tex]

       [tex]F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)[/tex]

       [tex]G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)[/tex]

       [tex]Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)[/tex]

2)

Answer:

4th ans

Explanation:

Answer:

join my z o o m

Explanation:

Answer:

A) a = 5.673 m/s²

The direction will be upwards vertically towards the point where it is suspended.

B) T = 113.2 N

Explanation:

A) We are given;

Weight of bowling ball; W = 71.7 N

Speed; v = 4.6 m/s

Rope length; r = 3.73 m

Now, formula for the centripetal acceleration is;

a = v²/r

Thus; a = 4.6²/3.73

a = 5.673 m/s²

The direction will be upwards vertically towards the point where it is suspended.

B) since weight is 71.7 N, it means that;

Mass = weight/acceleration = 71.7/9.8

Mass(m) = 7.316 kg

Thus,

Centripetal force is;

F_cent = 7.316 × 5.673

F_cent = 41.5 N

Thus, Tension in the rope is;

T = W + F_cent

T = 71.7 + 41.5

T = 113.2 N

Answer:

The magnitude of the force exerted by the person is 100 N

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn = ma

Where a is the acceleration of the object.

The net force is the sum of all forces exerted over a body. When an object is moved along a rough surface it experiences two horizontal forces and two vertical forces (provided there is no vertical component of the applied force).

The vertical forces are the Normal and the Weight and they are balanced, i.e.: N = W = mg.

The horizontal forces are The applied force (Fa) and the friction force (Fr). They are not balanced because the object is accelerated in that direction. The net force is:

Fn = Fa - Fr

Applying the first equation:

Fa - Fr = ma

Solving for Fa:

Fa = Fr + ma

Substituting the given values m=5 kg, Fr=10 N, [tex]a=18\ m/s^2[/tex].

Fa = 10 + 5*18 = 10 + 90 = 100

Fa = 100 N

The magnitude of the force exerted by the person is 100 N

Basically.

Given: F + m x a

Equation: 10 N + 5.0 kg x 18 m/s2

Solve: 100 N

Answer:

the rate at which his body is losing heat by conduction is 93.6 J/s

Explanation:

Given that;

surface area A = 1.8 m²

Skin temperature of the person Tp = 32°C = ( 34 + 273.15 ) = 307.15 K

Temperature of Air [tex]T_{air}[/tex] = 24°C = ( 24 + 273.15 ) = 297.15 K

Temperature of wall [tex]T_{wall}[/tex] = 17°C = ( 17 + 273.15 ) = 290.15 K

Length ( thick dead layer = 5 mm = 0.005 m

Skin emissivity = 0.97

Rate of metabolism = 155 W

rate his body is losing heat by conduction = ?

first we determine the difference in temperature between the skin and air

so

ΔT = 307.15 K - 297.15 K = 10 K

we know that; coefficient of thermal heat conductivity of air k = 0.026 W/mK

so

rate of heat loss by conduction Q/ΔT will be;

Q/ΔT = (KA/L)ΔT

so we substitute

= ( 0.026 × 1.8/ 0.005 )10

= 9.36 × 10

= 93.6 J/s

Therefore, the rate at which his body is losing heat by conduction is 93.6 J/s

Answer:

atmosphere

Explanation:

.........,....

Answer:

It will take 29.31 seconds for the Boxster to catch the Scion

Explanation:

Given the data in the question;

lets say Toyota Scion xB is car A and Porsche Boxster convertible is B and Toyota Scion xB is car A

the distance travelled by car A is

x = [tex]V_{A}[/tex] × t

where  [tex]V_{A}[/tex] is the speed of the car and t is time

the distance travelled by car B before reaching car A will be;

x + x₀ = [tex]V_{B}[/tex] × t

Now lets replace x by [tex]V_{A}[/tex] × t

so

([tex]V_{A}[/tex] × t) + x₀ = [tex]V_{B}[/tex] × t

x₀ = ([tex]V_{B}[/tex] × t) - ([tex]V_{A}[/tex] × t)

x₀ = t ([tex]V_{B}[/tex] - [tex]V_{A}[/tex])

t = x₀ /  ([tex]V_{B}[/tex] - [tex]V_{A}[/tex])

so we substitute

t = 170 m  /  (24.4 - 18.6)  

t = 170 / 5.8

t = 29.31 s

Therefore; it will take 29.31 s for the Boxster to catch the Scion

Answer:

Hey dear

Explanation:

Its option C

As Wavelength increases frequency decreases

In other case,

When Wavelength decreases frequency increases

its opposite

Tq

Answer:

64.2 m/s

Explanation:

We are given that

Speed ,v=38 m/s

We have to find the maximum speed when his car reach on flat ground.

Using dimensional analysis

[tex]F_{res}\propto v^2[/tex]

If 35% acceleration reduced by F(res) at 38 m/s

Then, 100% acceleration  can be reduced  by F(res) at v' m/s

[tex]\frac{F_1}{F_2}=\frac{v^2}{v'^2}[/tex]

[tex]v'^2=\frac{F_2}{F_1}v^2[/tex]

[tex]v'=v\sqrt{\frac{F_2}{F_1}}[/tex]

Substitute the values

[tex]v'=38\times \sqrt{\frac{100}{35}}[/tex]

[tex]v'=64.2 m/s[/tex]

Hence, the maximum speed when his car can reach on flat ground=64.2 m/s

The maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.

The air resistance is directly proportional to the square of the velocity of an object.

R ∝ v²

The speed of the car was reduced by 35 % at 38 m/s.

So, the speed of the car was reduced by 100% at v' m/s.

The relationship can be given by,

[tex]\dfrac {R_1}{R_2} = \dfrac {v^2}{v'^2}[/tex]

Put the values in the formula and calculate for [tex]v'[/tex],

[tex]v' = \sqrt {\dfrac {R_2 v^2}{R_1 }}[/tex]

[tex]v' = \sqrt {\dfrac { 100\times 38 ^2}{35 }}\\\\v' = 62.2 \rm \ m/s[/tex]

Therefore, the maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.

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Answer:

The raft moves towards the shore

Explanation:

The gravitational force exerted by the moon on the satellite is such that

F = G M m / R ² = m a   →   a = G M / R ²

where a is the satellite's centripetal acceleration, given by

a = v ² / R

The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that

v = 2πR / T   →   a = 4π ² R / T ²

Substitute this into the first equation and solve for T :

4π ² R / T ² = G M / R ²

4π ² R ³ = G M T ²

T ² = 4π ² R ³ / (G M )

T = √(4π ² R ³ / (G M ))

T = 2πR √(R / (G M ))

The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:

[tex] F = \frac{GmM}{R^{2}} [/tex]    (1)

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

R: is the distance between the satellite and the center of the moon

m: is the satellite's mass

M: is the moon's mass

The gravitational force is also equal to the centripetal force:

[tex] F = ma_{c} [/tex]   (2)

The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):

[tex] a_{c} = \frac{v^{2}}{R} [/tex]   (3)

And from the tangential velocity we can find the period:

[tex] v = \omega R = \frac{2\pi R}{T} [/tex]   (4)

Where:

ω: is the angular speed = 2π/T

By entering equations (4) and (3) into (2), we have:

[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex]   (5)

By equating (5) and (1), we get:

[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]

[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]  

[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]

[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]

Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

Find more here:

I hope it helps you!

Answer:

The magnitude of the work done by the worker is 303 J.

Explanation:

The work done by the worker can be found as follows:

[tex] W = |F|\cdot |d| cos(\alpha) [/tex]                              

Where:

F: is the force applied by the worker

d: is the displacement = 5.0 m

α: is the angle between the force applied and the displacement = 180°

We need to find the force applied by the worker:

[tex]\Sigma F = ma[/tex]

Taking as positive the movement direction of the crate we have:

[tex] -F - F_{\mu} + P_{x} = 0 [/tex]  

Where:              

m: is the crate's mass

a: is the acceleration = 0 (It is moving at constant speed)

F: is the force applied by the worker

Pₓ: is the weight in the horizontal direction    

[tex]F_{\mu}[/tex]: is the frictional force  

Hence, the force applied by the worker is:            

[tex]F = P_{x} - F_{\mu} = mgsin(\theta) - \mu mgcos(\theta)[/tex]

[tex] F = 50 kg*9.81 m/s^{2}*(sin(25) - 0.33cos(25)) = 60.6 N [/tex]  

Then, the work done by the worker is:                        

[tex] W = |F|\cdot |d| cos(\alpha) = 60.6 N*5.0 m*cos(180) = -303 J [/tex]                              

Therefore, the magnitude of the work done by the worker is 303 J.

I hope it helps you!                                            

The work is done by the worker will be 303 J. Work done is described as the multiplication of applied force and the amount of displacement.

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.

The given data in the problem is;

F is the force applied by the worker

d is the displacement = 5.0 m

α is the angle between the force applied and the displacement = 180°

m is the crate's mass=  50 kg

a is the acceleration = 0

Pₓ: is the weight in the horizontal direction    

The net force on the crate is found as;

[tex]\rm F_{net}= P_X - F_{\mu} \\\\ \rm F_{net}= mg sin \theta - \mu mg cos \theta \\\\ \rm F_{net}=50 \times \times 9.81 (sin 25^0 -0.33 cos(25) \\\\ \rm F_{net}= 60.6 N \\\\[/tex]

The work done  by the worker will be;

[tex]\rm W= Fd cos \alpha \\\\ \rm W= 60.6 \times 5.0 cos 180^0 \\\\\rm W=-303 \ J[/tex]

Hence the work is done by the worker will be 303 J.

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Answer:

F = 1176 N

Explanation:

Given that,

Mass of a baseball, m = 0.140-kg

Initial speed of the baseball, u = 33.6 m/s

Final speed of the baseball, v = -33.6 m/s (in opposite direction)

The time of contact of the bat and the ball, t = 0.008 s

We need to find the  average force the bat exerts on the ball. The force acting on the ball is given by :

[tex]F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.14(-33.6-33.6)}{0.008}\\F= -1176\ N[/tex]

So, the average force the bat exerts on the ball is 1176 N.

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

so

m = 4000 / 2.20462 =  1814.37 kg

Initial velocity [tex]V_{i}[/tex] = 60 mph = 26.8224 m/s

Final velocity [tex]V_{f}[/tex] = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = [tex]\frac{1}{2}[/tex]m(  [tex]V_{i}[/tex]² - [tex]V_{f}[/tex]² )

we substitute

Δk = [tex]\frac{1}{2}[/tex]×1814.37( (26.8224)² - (13.4112)² )

Δk = [tex]\frac{1}{2}[/tex] × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

so

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Answer:

 the statement is False       [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ

Explanation:

Let's analyze this problem, the ball and the receiver leave the same point and we want to know if at the same moment they reach the same point, for this we must have both the ball and the receiver travel the same distance.

Let's start by finding the time it takes for the ball to reach the ground

        y = [tex]v_{oy}[/tex] t - ½ g t²

when it reaches the ground its height is y = 0

       0 = vo sin θ  - ½ g t²

       0 = t (vo sin θ - ½ g t)

The results are

       t = 0                             exit point

       t = 2 v₀ sin θ/g            arrival point

at this point the ball traveled

       [tex]x_{ball}[/tex]= v₀ₓ t

       x_{ball} = v₀ cos θ  2v₀ sin θ / g

       x_{ball}= 2 v₀² cos θ   sin θ/ g

Now let's find that distantica traveled the receiver in time

        [tex]x_{rec}[/tex] = v₀ t

        x_{rec} = v₀ (2 v₀ sin θ / g)

        x_{rec} = 2 v₀² sin θ / g

without dividing this into two distances

          [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ

therefore the distances are not equal to the ball as long as behind the receiver

Therefore the statement is False

Answer:

a)

Explanation:

        [tex]x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)[/tex]

       [tex]a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)[/tex]

       [tex]x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)[/tex]

       [tex]a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)[/tex]

       [tex]x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)[/tex]

Answer: The correct option is A.

Move toward X but not rotate

Explanation:

This is because from the question, X is fixed and y is free to move. Since magnetic dipole of X is fixed, that is it can't move and that of y is free to move, therefore y will move toward x because it's forces of attraction is linear and not rotational and besides X and Y are on a linear path, therefore Y will move towards X that is fixed and it will not rotate since it's linear.

Answer:

Risk rejection

Explanation:

There are several factors that contribute to the degree of driving risks and they include but not limited to the ability of the driver and the condition of a vehicle. Other factors are condition of the environment and the condition of the highway. When driving, a driver may wait until an oncoming vehicle passes before making a complete left turn as a risk rejection strategy. Left turns are more dangerous when making them because drivers tend to accelerate on to a left turn. The wider radius of a  left turn is know to led to higher speeds and greater pedestrian exposure. A driver is advised to have more mental and physical efforts when making a left turn.

Answer:

1.5m/s

Explanation:

Given parameters:

M1  = 75kg

V1  = 4m/s

M2  = 125kg

V2  = 0m/s

Unknown:

Final velocity of the two objects  = ?

Solution:

To solve this problem, we must understand that this is an inelastic collision. To conserve momentum;

          M1 V1   + M2 V2  = V (M1 + M2)

       ( 75 x 4 )  +  ( 125 x 0)   = V (75 + 125)

                   300  = 200V

                     V  = 1.5m/s

Answer:

The chemical formula for zinc (III) phosphide is Zn3P2

Answer: Out of the 5, the first 3 that came to my mind first were thermal energy, electrical energy, and sound energy.

Explanation:

Thermal-Generates due to the quick motion of atoms and molecules, especially when they collide with each other. It is also called heat energy. The matters in the universe consist of atoms or molecules which are always in motion. However, we can’t see the movement of this energy with our naked eyes. We can feel it at the time it touches our skin.

Sound-The vibration of an object causes sound energy. Sound is the movement of energy generated by vibrations through some substance, such as air or water or solid. Sound energy can travel through any medium to transfer energy from one particle to another, and you can hear it. However, it cannot travel through a vacuum as a vacuum does not contain any particles that can act as a medium. When an object vibrates, it makes the surrounding particles vibrate by transferring its energy. These particles, when colliding with other particles, make them vibrate. In this way, the sound energy gets transferred from one particle to another.

Electrical- Every object in the universe is made up of small particles called atoms. Atoms consist of tiny particles, namely electrons, protons, and neutrons. The electrons in the atom always move around the nucleus of an atom. While applying voltage, the electrons present in the atom get energy and break the bonding with the parent atom and thus become free. The energy of this free electron is called electrical energy or electricity. Therefore, it is the energy of these moving free electrons. Electrons are negatively and positively charged and usually move through a wire.

Answer:

The answer is below

Explanation:

Firstly, the frequency is received by the wall and then it is reflected and received by the bat.

The frequency received by the wall (f') is given by:

f' = [tex]f(v\pm v_o)/v\\\\[/tex]

The - sign is used when the  observer is moving away from the source and + sign when the observer is moving towards the source.

Since the bat is moving towards the wall, we use a positive sign. Hence:

f' = [tex]f(v+ v_o)/v\\\\[/tex]

The frequency reflected and received by the bat f" is:

f'' = [tex]f'\frac{v}{ (v\pm v_s)}\\\\[/tex]

- sign is used when the source moves toward the observer and + is used when the source moves away

since the bat moves towards the wall, then::

f'' = [tex]f'\frac{v}{(v-v_s)} =\frac{f(v+v_o)}{v}*\frac{v}{(v-v_s)} =f\frac{(v+v_o)}{(v-v_s)} \\\\[/tex]

v = speed of sound in air = 331 m/s, vo = velocity of observer = 5 m/s, vs = velocity of source = 5 m/s. Therefore:

[tex]f"=f\frac{(v+v_o)}{(v-v_s)} =40\ kHz\frac{(331\ m/s+5\ m/s)}{(331\ m/s+5\ m/s)} \\\\f"=41\ kHz[/tex]

The frequency of the echo received by the bat is 38.8 kHz.

The given parameters:

The observed frequency or frequency received by the bat is calculated by applying Doppler effect as follows;

[tex]f_0 = f_s(\frac{v \ +/-\ v_0}{v\ +/- \ v_s} )[/tex]

Since the bat is flying away from the wall the frequency received will be smaller than the actual frequency;

[tex]f_o = f_s (\frac{v \ - \ v_0}{v \ + \ v_s} )\\\\f_o = 40 \ kHz \times (\frac{331 - 5}{331 + 5} ) \\\\f_0 = 38.8 \ kHz[/tex]

Thus, the frequency of the echo received by the bat is 38.8 kHz.

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