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2 years ago

Which of the following is MOST useful to scientists in measuring the size of asteroids?

Physics

1 answer:

Alenkasestr [34]2 years ago

5 0

Answer:c-The gravitational effect when spacecraft flies close to the asteriod

Explanation:

Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.

The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.

Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.

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find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds

Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, $u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}$

Final speed of the train, $v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}$

Displacement of the train, $S=234\textrm{ km}=234\times 1000=234000\textrm{ m}$

Using Newton's equation of motion,

$v - u = at\\a=\frac{v-u}{t}$

Now, using Newton's equation of motion for displacement,

$v^{2}-u^{2}=2aS$

Now, plug in the value of $a=\frac{v-u}{t}$ in the above equation. This gives,

$v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}$

Now, plug in 234000 m for $S$ , 25 m/s for $v$ and 20 m/s for $u$ . Solve for $t$ .

$t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}$

Therefore, the time taken by the train is 10400 s.

3 0

3 years ago

5. The volume of physical activity attained by an individual who exercises at a level of 7 METs for 35 min · day-1, 4 days · wk-

Airida [17]

D. 980, this is the best answer because 35 x 7 is 980 :)

3 0

2 years ago

To solve a problem using the equation for keplerâs third law, enrico must convert the average distance of mars from the sun from

Natali5045456 [20]

1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.

3 0

3 years ago

Read 2 more answers

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

$\omega=-\alpha t$

$\alpha=-\dfrac{\omega}{t}$

$\alpha=-\dfrac{50.0}{20.0}$

$\alpha=-2.5\ rad/s^2$

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

$\theta=\omega_{0}t+\dfrac{1}{2}\alpha t$

Put the value into the formula

$\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2$

$\theta=500\ rad$

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

$\vec{\tau}=\vec{r}\times\vec{f}$

$\tau=r\times f\sin\theta$

Put the value into the formula

$\tau=2.5\times4\times 9.8\sin60$

$\tau=84.87\ N-m$

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0

3 years ago

If a 5 N force pushes a 20 kg mass on a frictionless surface, how fast is the mass going in 5 seconds.

stira [4]

Explanation:

Let me know if you have questions

3 0

3 years ago