Ask question

Ask question

All categories

3 years ago

15

8. A book is pushed a distance of 0.78 m by a force that gives the book an acceleration of 1.54 m/s². If 1.56 J of work is done

on the book, what is the book’s mass?

1 answer:

kozerog [31]3 years ago

6 0

Answer:

1.29 kg

Explanation:

As we know that

W= FS

1.56 = F (0.78)

F= 1.56/0.78

F= 2 N

Now, According to the definition of Force

F= ma

then

m= F /a

m= 2 / 1.54

m = 1.29 kg

You might be interested in

The chart shows data for a moving object.

Ipatiy [6.2K]

Answer:

number 2

Explanation:

8 0

3 years ago

Which is a major threat to biodiversity? A. Red-billed oxpeckers eat ticks off of impalas' coats in the grasslands of East Afric

SIZIF [17.4K]

B. Mining in the Guinean Forests of West Africa to provide diamond and gold jewelry for humans.

4 0

4 years ago

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

3 years ago

A magnetic field would be produced by a beam of

kumpel [21]

3 protons should be your answer

4 0

3 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th

timama [110]

Answer:

Explanation:

Given

length of window $h=2.9\ m$

time Frame for which rock can be seen is $\Delta t=0.134\ s$

Suppose h is height above which rock is dropped

Time taken to cover $h+2.9 is t_1$

so using equation of motion

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h is

$h=0+0.5\times g\times (t_2)^2---2$

Subtract 1 and 2 we get

$2.9=0.5g(t_1^2-t_2^2)$

$5.8=g(t_1+t_2)(t_1-t_2))$

and from equation $t_1-t_2=0.134\ s$

so $t_1+t_2=\frac{5.8}{9.8\times 0.134}$

$t_1+t_2=4.416\ s$

and $t_1=t_2+\Delta t$

so $t_2+\Delta t+t_2=4.416$

$2t_2+0.134=4.416$

$t_2=0.5\times 4.282$

$t_2=2.141\ s$

substitute the value of $t_2$ in equation 2

$h=0.5\times 9.8\times (2.141)^2$

$h=22.46\ m$

8 0

3 years ago