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3 years ago

A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall? ( g = 10m/s2 )

Physics

1 answer:

LekaFEV [45]3 years ago

4 0

Answer:

Height is 11.25m

Explanation:

<u>Given the following data;</u>

Initial velocity, u = 0

Final velocity, v = 15m/s

Acceleration due to gravity, g = 10m/s²

To find the height, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement (height) measured in meters.

$V^{2} = U^{2} + 2aS$

<em>Making S the subject, we have;</em>

$S = \frac {V^{2} - U^{2}}{2a}$

But a = g = 10m/s²

<em>Substituting into the equation, we have;</em>

$S = \frac {15^{2} - 0^{2}}{2*10}$

$S = \frac {225 - 0}{20}$

$S = \frac {225}{20}$

S = 11.25m

<em>Therefore, the ball will reach a height of 11.25m before it begins to fall. </em>

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Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

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3 years ago

A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffici

Ivenika [448]

Answer:

Coefficient of friction will be 0.296

Explanation:

We have given initial speed of the stone u = 8 m /sec

It comes to rest so final speed v = 0 m /sec

Distance traveled before coming to rest s = 11 m

According to third equation of motion

$v^2=u^2+2as$

So $0^2=8^2+2\times a\times 11$

$a=\frac{-64}{22}=-2.90m/sec^2$

Acceleration due to gravity $g=9.8m/sec^2$

We know that acceleration is given by

$a=\mu g$

So $2.90=9.8\times \mu \\$

$\mu =\frac{2.9}{9.8}=0.296$

So coefficient of friction will be 0.296

3 0

3 years ago

A bee flies at a constant rate of 5 m s until it notices a flower. It then comes to rest in 5 seconds. Find the acceleration of

FrozenT [24]

Answer:

B) -1m/s^2

Explanation:

Final speed = 0 m/s

Initial speed = 5m/s

Time taken for it to come to rest(0m/s) = 5

then use the formula;

[v = u + at],where v is the final speed..u is the initial speed..t is time taken for it to come to rest and a is the acceleration

; 0 = 5 + 5a

; -5 = 5a

;Acceleration = -1 m/s^2

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3 years ago

A pendulum of length =1.0 m is pulled to the side and released on the moon. It's period is measured to be 4.82 seconds. What is

Shkiper50 [21]

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

It is given that,

Length of pendulum, l = 1 m

Time period, T = 4.82 seconds

We have to find the gravity of the moon. The time period of the pendulum is given by :

$T=2\pi\sqrt{\dfrac{l}{g}}$

g = acceleration due to gravity on moon

$g=\dfrac{4\pi^2l}{T^2}$

$g=\dfrac{4\pi^2\times 1\ m}{(4.82\ s)^2}$

g = 1.69 m/s²

Hence, the gravity on the moon is 1.69 m/s².

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4 years ago

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A centripetal force of 210 N acts on a 1,600-kg satellite moving with a speed of 5,500 m/s in a circular orbit around a planet.

Dafna1 [17]

Answer:

Explanation:

Step one:

given

Force F= 210N

mass m= 1600kg

velocity v=5500m/s

Step two

Required is the radius r

the expression for the force is

$F_c = \frac{mv^2}{r}$

substitute

210=1600*5500^2/r

cross multiply we have

210r=48400000000

divide both side by 210

r=230476190.476m

r=230476.19km

4 0

3 years ago