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aalyn [17]
3 years ago
11

If a ball rolls off a 4m high table at 7m/s, how long will it take until it hits the floor

Physics
1 answer:
Nat2105 [25]3 years ago
3 0

Answer:

0.9 s

Explanation:

Given (vertical direction):

y₀ = 4 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (4 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t ≈ 0.9 s

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What is alpha, beta and gamma radiation made of
algol13

Answer:

Alpha radiation is the name for the emission of an alpha particle in fact an helium nuclei, beta radiation is the emission of electrons or positrons , and gamma radiation is the term used for the emission of energetic photons.

Explanation:

6 0
1 year ago
If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th
mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

x=ut

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

y=\frac{1}{2} gt^2

Substitute 9.8 m/s² for g and 0.461 s for t.

y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,(2.0 m)-(1.04 m)=0.96 m

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0
3 years ago
A 3 kg object is attached to a 1000 N/m spring. The spring is compressed 0.10 m and then the spring launches the object horizont
Viktor [21]

Answer:

1.826m/s

Explanation:

E=1/2*k*(∆L)^2

E=1/2*mV^2

6 0
3 years ago
Which of the following statements best describes an electromagnetic wave with a long wavelength?
zmey [24]
It has a high frenquency and can only travel through a medium
5 0
4 years ago
If the distance between two asteroids is halved, the gravitational force they exert on each other will
mamaluj [8]

Answer:

e) Be four times greater

Explanation:

Here we have to use Newton's gravitational law that relates the gravitational force between two objects with their masses (m_{1} & m_{2}) and the distance between them (r) in the next way:

F=G\frac{m_{1}m_{2}}{r^{2}} (2)

Now if distance between asteroids is halved:

F_{2}=G\frac{m_{1}m_{2}}{(\frac{r}{2})^{2}}

F_{2}=G\frac{m_{1}m_{2}}{\frac{r^{2}}{4}}

F_{2}=G\frac{4m_{1}m_{2}}{r^{2}}=4G\frac{m_{1}m_{2}}{r^{2}}

Note that G\frac{m_{1}m_{2}}{r^{2}} because (1) is F so:

F_{2}=4F

It's four times greater!

3 0
3 years ago
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