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asambeis [7]
3 years ago
8

A pencil is rolled off a table of height 0.92 m. If it has horizontal speed pf 1.4 m/s, how long does it take the pencil to reac

h the ground?
Physics
1 answer:
Alenkasestr [34]3 years ago
3 0

The distance an object falls, from rest, in gravity is

                         D  =  (1/2) (G) (T²)

                        'T' is the number seconds it falls.

In this problem,

                         0.92 meter = (1/2) (9.8) (T²)

Divide each side by  4.9 :   0.92 / 4.9 = T²

Take the square root
of each side:                          √(0.92/4.9) = T

                                                  0.433 sec = T    

The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor.  BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.

The pencil is in the air for  0.433 second.
In that time, it travels
                                   (0.433s) x (1.4 m/s) = 0.606 meter

from the edge of the table.
 
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3 years ago
Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
3 years ago
A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin
emmainna [20.7K]

(1)

Cheetah speed: v_c = 97.8 km/h=27.2 m/s

Its position at time t is given by

S_c (t)= v_c t (1)

Gazelle speed: v_g = 78.2 km/h=21.7 m/s

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

S_g(t)=S_0 +v_g t (2)

The cheetah reaches the gazelle when S_c=S_g. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

v_c t=S_0 + v_g t

(v_c -v_g t)=S_0

t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s


(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: S_c = v_c t =(27.2 m/s)(7.5 s)=204 m

Gazelle: S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m

So, the gazelle should be ahead of the cheetah of at least

d=S_c -S_g =204 m-162.8 m=41.2 m

4 0
3 years ago
Read 2 more answers
A 438kg car is accelerating east at 2.55m/s^2. What is the total force acting east on the car
lisabon 2012 [21]

Answer:

<h2>1116.9 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 438 × 2.55

We have the final answer as

<h3>1116.9 N</h3>

Hope this helps you

5 0
3 years ago
Which element of variation would not be affected by adding the data value 15 to the data set {3, 5, 6, 8, 9, 10, 13, 14}?
Anna11 [10]
The answer to this would be 15
4 0
3 years ago
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