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2 years ago

A golf ball is hit by a club. The graph shows the variation with time of the force exerted on the bal

Physics

2 answers:

sammy [17]2 years ago

6 0

Answer:

D the maximum acceleration of the ball

Explanation:

A the average force on the ball

Total change in momentum of the ball is area under the graph between force and time

So here average force is the ratio of area under the force time graph and total time of the graph

B the change in momentum of the ball

Total change in momentum of the ball is area under the graph between force and time

C the contact time between the ball and the club

Here total time of contact is the total time for which force is acting on the ball due to wall

D the maximum acceleration of the ball

Here since we do not know about the mass of ball we can not find the acceleration.

So here we have correct answer as

D the maximum acceleration of the ball

emmasim [6.3K]2 years ago

5 0

You can't find the acceleration of the ball. The graph tells the force, but you'd also need to know the mass of the ball.

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A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

$F = \dfrac{mv^2}{r}$

m is the mass, v is the velocity and r is the radius.

It follows that $F \propto v^2$ , provided m and r are constant.

When v is doubled, the new force, $F_1$ , is

$F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F$

Hence, the tension in the string is quadrupled.

8 0

2 years ago

Tori experiments with pulleys in physics class. She applies 70 newtons of force to a single pulley to lift a bowling ball. By ad

e-lub [12.9K]

The minimum input force she'll need to lift the ball is 35 N.

Explanation:

Mechanical advantage of a single pulley is 1. As, she applies 70 N of force to lift the bowling ball, so the output force(weight of the ball) is also 70 N.

Now, adding another pulley gives a mechanical advantage of 2. We have,

M.A = (Output Force)/(Input Force)

Substituting the values we get,

$2=\frac{70}{F_{i}}$

$F_{i}=\frac{70}{2}$ = 35 N

Input force equals to 35 N needs to be applied.

6 0

2 years ago

Read 2 more answers

A cyclist moving with a constant velocity of 6.0 m/s forward passes a car that is just starting. If the car has a constant accel

sergiy2304 [10]

After 6 seconds, the car will surpass the cyclist.

<h3><u>Explanation:</u></h3>

The speed of the cyclist = 6 m/s.

Let after time t sec, the car will overtake the cyclist.

So, distance covered by the cyclist in t sec = 6t m

Initial velocity of the car is 0 m/s, because the car is just starting.

Acceleration of the car = $2 m/s^2$ .

Final velocity of the car =6 m/s.

So to cover the distance 6t, the time required by the car = $\frac{1}{2} \times a \times t^2 = \frac{1}{2} \times 2 \times t^2$

$6t = \frac{1}{2} \times 2 \times t^2\\6t = t^2$

t =6 sec

So, after 6 seconds, the car will surpass the cycle.

6 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago

If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

The mass of the rocket is m = 6300 kg

The velocity at gas is being ejected is u = 2000 m/s

The initial acceleration desired is $a = 27.0 \ m/s$

The time taken for the gas to be ejected is t = 1 s

Generally this desired acceleration is mathematically represented as

$a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$ is the rate at which gas is being ejected with respect to time

Substituting values

$27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=> $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=> $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=> $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=> $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

3 0

2 years ago