Rutherford proposed that

Two observers times the motion of a car from one place to another. The first observer's clock read 262 seconds at the start and

icang [17]

Answer:

The time of the second observer at the end of the car motion is 27 s

Explanation:

Initial time of the first observer, t₁ = 262 s

final time of the first observer, t₂ = 375 s

The time of the car motion, t = t₂ - t₁

t = 375 s - 262 s

t = 113 s

Initial time of the second observer, t₁ = -86 s

final time of the second observer, t₂ = ?

The time of the car motion, t = t₂ - t₁

113 = t₂ - t₁

113 = t₂ - (-86)

113 = t₂ + 86

t₂ = 113 - 86

t₂ = 27 s

Therefore, the time of the second observer at the end of the car motion is 27 s

4 0

4 years ago

Suppose a spacecraft orbits the moon in a very low, circular orbit, just a few hundred meters above the lunar surface. The moon

Finger [1]

Spacecraft will be moving in 1700 m/s.

Option C

<h3>Explanation: </h3>

The diameter of the moon is 3500 km and the free fall acceleration at the surface is given as $1.6\ \mathrm{m} / \mathrm{s}^2$

The radius will be half of the diameter of the moon that can be written as:

$r_{\text {moon }}=1.75 \times 10^{6}$

By the application of the equation for orbit speed, we get

$\begin{aligned}&v_{\text {orbit}}=\sqrt{r \times q}\\&v_{\text {orbit}}=\sqrt{\left(1.75 \times 10^{6}\right) \times\left(1.6\ \mathrm{m} / \mathrm{s}^{2}\right)}\\&v_{\text {orbit}}=1700\ \mathrm{m} / \mathrm{s}\end{aligned}$

5 0

3 years ago

Camping equipment weighting 5000N is pulled across a frozen lake by means of a horizontal rope. There is a frictional force of 3

Dmitry [639]

Answer:

The work done by the campers is $4\times10^{5}\ J$

(b) is correct option.

Explanation:

Given that,

Weight = 5000 N

Frictional force = 300 n

Distance = 1000 m

Constant rate of speed = 0.20 m/s²

We need to calculate the force

Using newton's law of motion

$F-F_{\mu}=ma$

$F-300=\dfrac{5000}{10}\times0.20$

$F=\dfrac{5000}{10}\times0.20+300$

$F=400\ N$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=400\times1000$

$W=4\times10^{5}\ J$

Hence, The work done by the campers is $4\times10^{5}\ J$

3 0

4 years ago

Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller

IgorC [24]

Answer:

Explanation:

i )

In a conservative field like gravitational field , loss of potential energy or work done , depends upon the initial and final point and not the manner in which 2 nd point has been reached . Since the initial and final point is same in both the cases of straight and curved path , final velocity will remain same for both of them .

Hence , due to increased mass of larger child , his kinetic energy will be greater .

ii ) Since the initial and final point is same in both the cases of straight and curved path , final velocity will remain same for both of them .

iii ) Smaller child undergo free fall , therefore , he will fall with acceleration g . The larger child falls on curved path . So , he will have only a component of

vertical g at any moment . hence average acceleration will be less.

6 0

4 years ago

A team of dogs pulls a sled across the snow. The weight of the loaded sled is 2,200 N. The coefficient of friction is 0.15. How

chubhunter [2.5K]

.. The weight of the loaded sled is 2200 N. The coefficient of friction is 0.15. How hard must the dogs pull so that their effort just equals the frictional force? ... to move after the light changes from red to green A team of dogs pulls a sled across the snow. ... 2) pulling force of dogs fL = us N = 0.15 x2200 = 330 N

DOG:
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7 0

4 years ago

Read 2 more answers