Ask question

Ask question

All categories

3 years ago

8

A 730 N student stands in the middle of a frozen pond having a radius of 5.8 m. He is unable to get to the other side because of

a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 2.6 kg physics textbook horizontally toward the north shore at a speed of 6.1 m/s. The acceleration of gravity is 9.81 m/s². How long does it take him to reach the south shore? Answer in units.

1 answer:

dedylja [7]3 years ago

7 0

Answer:27.35 s

Explanation:

Given

weight of man =730 N

mass of man $m_1=\frac{730}{9.8}=74.48 kg$

radius of pound r=5.8 m

mass of book $m_2=2.6 kg$

velocity of book $v_2=6.1 m/s$

let $v_1$ be the velocity of man

conserving momentum

$0=m_1v_1+m_2v_2$

$0=74.48\times v_1+2.6\times 6.1$

$v_1=\frac{15.86}{74.48}=0.212 m/s$

time taken by man to reach south end

$t=\frac{5.8}{0.212}=27.35 s$

You might be interested in

Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message e

Grace [21]

Answer:

The duration is $T =72 \ years /tex]Explanation:From the question we are told that The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m$

Generally the time it would take for the message to get the the other civilization is mathematically represented as

$t = \frac{D}{c}$

Here c is the speed of light with the value $c = 3.08 *10^{8} \ m/s$

=> $t = \frac{3.311 *10^{17} }{3.08 *10^{8}}$

=> $t = 1.075 *10^9 \ s$

converting to years

$t = 1.075 *10^9 * 3.17 *10^{-8}$

$t = 1.075 *10^9 * 3.17 *10^{-8}$

$t = 34 \ years$

Now the total time taken is mathematically represented as

$T = 2* t + 2 + 2$

=> $T = 2* 34 + 2 + 2$

=> [tex]T =72 \ years /tex]

4 0

2 years ago

A motorcyclist accelerates from rest to 10 mi/hr. what is the change in velocity

gulaghasi [49]

The change in velocity is 10 mi/h (4.47 m/s)

Explanation:

The change in velocity of the motorcyclist is given by

$\Delta v = v-u$

where

v is the final velocity

u is the initial velocity

In this problem, we have

u = 0 (the motorbike starts from rest)

v = 10 mi/h

Therefore, the change in velocity is

$\Delta v = 10 -0 = 10 mi/h$

And keeping in mind that

1 mile = 1609 m

1 h = 3600 s

We can convert it into m/s:

$\Delta v = 10 \frac{mi}{h} \cdot \frac{1609 m/mi}{3600 s/h}=4.47 m/s$

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

6 0

2 years ago

A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20

Klio2033 [76]

Answer:

$\boxed{{\boxed{\blue{ 12.5}}}}$

Explanation:

Given, for girl : Weight or force;

$\rm \: F_1 = 50 \: kgf$

Area of both heels;

$\rm \: A_1 = \; 2 ×1 \; cm^2 = 2 \: cm^2$

$\rm \: Pressure \: P_1 = \cfrac{F_1}{ A_1 } = \dfrac{50 \: kgf}{2 \: cm {}^{2} } = 25 \: kgf \: cm {}^{ - 1}$

For elephant, Weight = Force $\rm F_2$ = 2000 kg•f

Area of 4 feet;

$\rm \: A_2 = \; 4 \times 250 \; cm^2 = 1000 \: cm^2$

$\rm \: Pressure \: P_2 = {F_2}/{A_2} \; = \cfrac{2 \cancel{0 00 }\: kgf}{1 \cancel{000} \: cm^2} = 2 \: kgf \: { \:cm}^{- 1}$

Now;

$\rm = \dfrac{Pressure \: Exerted \: by \: the \: Girl}{Pressure \: exerted \: by \: the \: elephant}$

$= \rm \: P_1/P_2$

$\implies \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } = \rm\cfrac{25 \: \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}$

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0

2 years ago

4. A chandelier brightens a ballroom after a waiter moves a switch

Leokris [45]

Answer:

because switch moves after a waiter

Explanation:

7 0

3 years ago

If you were in charge of designing a wire to carry electricity across your city, state or province, which of

Anon25 [30]

Answer:

Thin, aluminium and buried underground.

Explanation:

When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.

i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.

ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.

iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.

7 0

2 years ago