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Lilit
21 hour ago
10

You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m/s. In stopping the ball, your hands and the volleyball

descend together a distance of 0.150 m.
How much work do your hands do on the volleyball in the process of stopping it?
What is the magnitude of the force (assumed constant) that your hands exert on the volleyball?
Physics
1 answer:
Softa [2K]21 hour ago
3 0

Explanation:

The amount of work performed corresponds to the variation in energy.

W = ΔKE

W = 0 − ½mv²

W = -½ (0.270 kg) (-7.50 m/s)²

W = -7.59 J

Work can be defined as the product of force and distance.

W = Fd

-7.59 J = F (-0.150 m)

F = 50.6 N

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A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir
ValentinkaMS [2425]

Answer:

The object measures 6 m in distance and 2 m in height.

It creates a virtual image that is upright.

Explanation:

Provided data includes:

Focal length = 0.25 m

Image height = 0.080 m

Image distance = 0.24 m

We are to determine the object's distance.

Using the lens formula:

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

Substituting values into the formula:

\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}

\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}

\dfrac{1}{u}=\dfrac{1}{6}

u=6\ m

We also need to calculate magnification:

Applying the magnification formula:

m=-\dfrac{v}{u}

Substituting values into this formula:

m=-\dfrac{0.24}{-6}

m=0.04

Next, we need to find the height of the object:

Using the magnification formula once more:

m=\dfrac{h'}{h}

h=\dfrac{0.080}{0.04}

h=2\ m

A convex mirror generates a virtual and upright image on its backside.

Consequently, the object is at a distance of 6 m and has a height of 2 m.

The image formed is virtual and upright.

6 0
26 days ago
Read 2 more answers
The magnitude of the force associated with the gravitational field is constant and has a value FF . A particle is launched from
serg [2593]

Answer:

The particle's kinetic energy will equal 12U₀.

Explanation:

Given that,

A particle is fired from point B with an initial speed and arrives at point A with U₀ joules of kinetic energy gained.

The consistent force acting is 12F.

As per the problem,

The kinetic energy is

U_{0}=Fx....(I)

Constant force remains at 12F.

A resistive force field now exists,

With the resistive force defined as,

F_{r}=12F

As the particle travels from point B to point A,

We need to find the kinetic energy

Using the kinetic energy formula

U=F_{r}x

Substituting values for F_{r}

U=12Fx

Then, from equation (I)

U=12U_{0}

Consequently, the particle’s kinetic energy will amount to 12U₀.

7 0
26 days ago
In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat
Yuliya22 [2420]

Answer:

-13.18°C

Explanation:

To solve this issue, we must examine the principles associated with the rate of thermal conduction.

This rate is defined by the equation

\frac{Q}{t} = \frac{kA\Delta T}{d}

Where,

Q = Amount of heat transferred

t = time

k = Thermal conductivity constant

A = Area of cross-section

\Delta T = Temperature difference across the material

d = Material thickness

The scenario indicates a heat loss that is double the initial value, which means

Q_2 = 2*Q_1

Substituting values yields,

kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}

\frac{\Delta T}{x}=2*\frac{\Delta T}{x}

\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}

\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}

Solving for T_o,

T_o = -13.18

Thus, when the heat lost per second is doubled, the temperature on the external surface of the window is -13.18°C.

3 0
1 month ago
Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
kicyunya [2264]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance for a metal can be calculated by

        R = ρ L / A

Where ρ indicates the resistivity of aluminum, L is the resistance's length, and A indicates the cross-sectional area

We use this formula for both configurations

For small face measurements (W x W)

The length is

         L = W

Area  

         A = W W = W²

         R₁ = ρ W / W² = ρ / W

For larger face measurements (D x L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

From this, we find the relation to be

    R₂ / R₁ = 2W²/L

6 0
1 day ago
Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee
Ostrovityanka [2204]

Answer: The frequency is 1714.3 Hz

Explanation: The calculation is derived from the Doppler effect formula.

Since the source is approaching the observer, the observer's velocity is considered positive.

Refer to the attached document for the detailed derivation

3 0
1 month ago
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