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3 years ago

Question 23

Physics

1 answer:

bonufazy [111]3 years ago

3 0

If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,

с Stay the same

Explanation:

The Mass of an object or body does not affect the acceleration due to gravity in any kind of way.
Light weight objects accelerate more slowly than the heavy objects because when the forces other than the gravity also plays a major role.
Mass increases of a body when an object has higher velocity or the speed.
The greater the force of gravity, it would give a direct impact on the object's acceleration; thus considering only a force, the heavier the object is, it would accelerate faster. But an acceleration depends upon the two factors which are force and mass.
Newton's second law of motion states that the acceleration of an object is dependent upon the two factors which are, the net force of an object and the mass of the object.

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All of the following are kinds of forces except

dangina [55]

Answer:

All of the following are kinds of forces except?

<h2><u><em>b) acceleration is the Answer</em></u></h2>

Explanation:

<h3><u><em>Acceleration is actually produced when a force acts upon a body therefore it is the result after we apply any force.</em></u></h3>

3 0

3 years ago

6.7.1 Current and Current Density A sphere of radius 10 mm that carries a charge of 8 nC =8x 10°C is whirled in a circle at the

aev [14]

Answer:

Part a)

Rate of charge flow is known as electric current

Part b)

Average current flow is

$i = 4\times 10^{-7} A$

Part c)

As we know that current density is current per unit area

$j = 1.27 \times 10^{-3} A/m^2$

Explanation:

As we know that angular frequency of rotation is

$\omega = 100\pi rad/s$

now by basic definition of electric current

Part a)

Rate of charge flow is known as electric current

$i = \frac{dq}{dt}$

so here we have

$i = \frac{Q}{T}$

$i = Qf$

Part b)

here we know that

$\omega = 2\pi f$

$100\pi = 2\pi f$

$f = 50 Hz$

now we have

$i = (8\times 10^{-9})(50)$

$i = 4\times 10^{-7} A$

Part c)

As we know that current density is current per unit area

So we have

$j = \frac{i}{A}$

$j = \frac{4\times 10^{-7}}{\pi(10\times 10^{-3})^2}$

$j = 1.27 \times 10^{-3} A/m^2$

6 0

3 years ago

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

sweet-ann [11.9K]

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

$\Delta T$ is the change in temperature

In this problem:

m = 1 g is the mass of water

$C=1 cal/g^{\circ}C$ is the specific heat capacity of water

$\Delta T=1^{\circ}C$ is the change in temperature

So, the heat needed is

$Q=(1)(1)(1)=1 cal$

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: $m=1g$

- Specific latent heat of fusion of ice: $\lambda_f=80 cal/g$

So, the heat needed is

$Q=(1)(80)=80 cal$

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: $m=1g$

- Specific latent heat of vaporization of water: $\lambda_v=540 cal/g$

So, the heat needed is

$Q=(1)(540)=540 cal$

5 0

3 years ago

What is the minimum speed needed to fire a champagne cork a distance of 11m?

Crank

To start with solving this problem, let us assume a launch angle of 45 degrees since that gives out the maximum range for given initial speed. Also assuming that it was launched at ground level since no initial height was given. Using g = 9.8 m/s^2, the initial velocity is calculated using the formula:

(v sinθ)^2 = (v0 sinθ)^2 – 2 g d

where v is final velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m

Rearranging to find for v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>

8 0

3 years ago

A 1000-kg car traveling at 70 m/s takes 3 m to stop under full braking. the same car under similar road conditions, traveling at

azamat

We assume $a=const$ (acceleration is constant. We apply the equation
$v^2=v0^2+2as$ where s is the distance to stop $v=0(m/s)$ . We find the acceleration from this equation
$a=-v0^2/(2s)=-70^2/(2*3) =-816.7 (m/s^2)
$
We know the acceleration, thus we find the distance necesssary to stop when initial speed is $v=140 (m/s)$
$s=-v0^2/(2a) =140^2/(2*816.7)=12 (m)$

5 0

3 years ago