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3 years ago

a 10kg ball is thrown into the air. it is going 3m/s when thrown. How much potential energy will it have at the top?

Physics

1 answer:

Alexandra [31]3 years ago

7 0

Answer:

45 J

Explanation:

Assuming the level at which the ball is thrown upwards is the ground level,

We can use the equations of motion to obtain the maximum height covered by the ball and then calculate the potential energy

u = initial velocity of the ball = 3 m/s

h = y = vertical distance covered by the ball = ?

v = final velocity of the ball at the maximum height = 0 m/s

g = acceleration due to gravity = -9.8 m/s²

v² = u² + 2ay

0 = 3² + 2(-9.8)(y)

19.6y = 9

y = (9/19.6)

y = 0.459 m

The potential energy the ball will have at the top of its motion = mgh

mgh = (10)(9.8)(0.459) = 45 J

Hope this Helps!!!

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jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

8 0

3 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

In this case:

x0= 0
v0= 0 because the object is initially stationary
a= 6 $\frac{m}{s^{2} }$
t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 $\frac{m}{s^{2} }$ *(15s)²

Solving:

x=½*6 $\frac{m}{s^{2} }$ *(15s)²

x=½*6 $\frac{m}{s^{2} }$ *225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

5 0

2 years ago

How might a nation be affected if another nation did not follow agreed upon water rules

forsale [732]

The other nation will get mad at the other nation and they could start a war
................................................................

4 0

3 years ago

A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight

photoshop1234 [79]

Answer:

$t=1.9 sec$

Explanation:

From the question we are told that:

Height $h=28m$

Time $t=3s$

Generally the Newton's equation for Initial velocity upward is mathematically given by

$s=ut+\frtac{1}{2}at^2$

$28=3u-\frac{1}{2}*9.8*3^2$

$u=24.03m/s$

Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28

$v=\sqrt{24.03-2*9.8*28}$

$v=5.35m/s and -5.35m/s$

Generally the Newton's equation for time to reach initial velocity is mathematically given by

$v=u+at$

$5.35=24.03-9.8t$

$t=\frac{28.03-5.35}{9.8}$

$t=1.9 sec$

4 0

2 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is