All categories

3 years ago

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

Physics

1 answer:

zimovet [89]3 years ago

6 0

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

$F=m\times a$

$F=m\times (\dfrac{v-u}{t})$

$F=0.003\times (\dfrac{30}{0.06})$

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

You might be interested in

In a Young's double-slit experiment, two parallel slits with a slit separation of 0.165 mm are illuminated by light of wavelengt

RSB [31]

Answer:

the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m

Explanation:

Given the data in the question;

slit separation d = 0.165 mm = 0.165 × 10⁻³ m

wavelength λ = 560 nm = 560 × 10⁻⁹ m

distance between the screen and slits D = 4.05 m

now,

for fifth-order bright fringe path difference = mλ

where m is 5

so, the difference in path lengths from each of the slits will be;

Δr = mλ

we substitute

Δr = 5( 560 × 10⁻⁹ m )

Δr = 28 × 10⁻⁷ m

Therefore, the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m

8 0

3 years ago

A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when

choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

$F_t - mg = ma$

here since speed is constant or it is at rest

so we will have

$a = 0$

$F_t = mg$

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

so tension force is less than the weight of crate if it is accelerating downwards

4 0

3 years ago

Formation of hydrogen bonds requires hydrogen atoms and what else?

ohaa [14]

Answer:

either a Nitrogen atom, Oxygen atom, or a Flourine atom

Explanation:

The atom has to be more electronegative than hydrogen for the bond to form.

8 0

3 years ago

The following is current scientific evidence supporting the nebular theory on the formation of the solar system. the composition

andreyandreev [35.5K]

All planets orbit the sun in a plane, all the planets orbit the sun in the same direction, most of the planets rotate in the same direction. I'm not sure when and answer ends or begins on your question so you can choose from some of the answers I gave you.

6 0

3 years ago

Read 2 more answers

The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it

MrRissso [65]

Answer:

Explanation:

From the question we are told that

The moment of inertia is $I = 2.50 \ kg \cdot m^2$