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Jobisdone [24]
3 years ago
5

You toss a bowling ball straight up into the air with a speed of 2.1 m/s. How long does it take the bowling ball to reach its hi

ghest point?
Physics
1 answer:
lara [203]3 years ago
6 0

Time taken by the bowling ball to reach its highest point= 0.214 s

initial velocity= Vi=2.1 m/s

Final velocity= Vf=0 as the velocity at the highest point is zero.

acceleration= g= -9.8 m/s²

using the kinematic equation Vf= Vi + at

0= 2.1 + (-9.8)t

t= -2.1/-9.8

t=0.214 s

Thus the time taken by the bowling ball to reach its highest point is 0.214 s

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What are the different types of Energy transformations and Energy Transfers?
Harman [31]

Explanation:

Here are some of the ways that energy can change (transform) from one type to another:

The Sun transforms nuclear energy into heat and light energy.

Our bodies convert chemical energy in our food into mechanical energy for us to move.

An electric fan transforms electrical energy into kinetic energy.

7 0
3 years ago
How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?
astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of  0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J

So, the magnitude of work done in lifting the book is 6 joules.

7 0
3 years ago
When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

3 0
3 years ago
Suppose a thick nimbostratus cloud contains ice crystals and cloud droplets all about the same size. Which precipitation process
Ronch [10]

Answer:The most important process would be the ice crystal process

Explanation:

Ice crystal process also called Bergeron process requires numerous small water drops that are supercooled, which is a common feature in clouds between about 0° and -20°C or below, along with a small number of ice crystals. Therefore because the collision-coalescence process requires that cloud droplets be of varying size so that drops will fill at different speeds, the most important process would be the ice crystal process.

7 0
3 years ago
A girl and a boy are riding on a merry go round that is turning at a constant rate. The girl is near the outer edge, and the boy
galina1969 [7]

Answer:

The girl has greater tangential acceleration

Explanation:

The angular acceleration (\alpha) of the merry go round is equal to the rate of the change of the angular velocity, \omega:

\alpha = \frac{d\omega}{dt}

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.

The tangential acceleration instead is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the distance from the centre of the merry go round

Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

5 0
3 years ago
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