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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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NEED ANSWER ASAP WILL MARK BRAINLIEST

lukranit [14]

Speed=30 m/s - 1.5 m/s = 28.5 m/s forward

7 0

3 years ago

Roller coasters accelerates from initial speed of 6.0 M/S2 final speed of 70 M/S over four seconds. What is the acceleration?

choli [55]

Answer:

$a=16\ m/s^2$

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The formula to calculate the change of velocities is:

$v_f=v_o+at$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The roller coaster moves from vo=6 m/s to vf=70 m/s in t=4 seconds. To calculate the acceleration, solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{70-6}{4}=\frac{64}{4}$

$\boxed{a=16\ m/s^2}$

3 0

3 years ago

Newtons laws of motion

Viktor [21]

Answer:

Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.

Law 2. A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force.

Law 3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.

7 0

2 years ago

Pls help and thank u need asap!

aleksandr82 [10.1K]

Answer:

ill help if u help me?its 4am

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2 years ago

A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What

kati45 [8]

Explanation:

The given data is as follows.

radius (r) = 3.25 cm, $\alpha = 11.6 rad/s^{2}$

Now, we will calculate the tangential acceleration as follows.

$a_{tangential} = \alpha \times r$

Putting the given values into the above formula as follows.

$a_{tangential} = \alpha \times r$

= $11.6 rad/s^{2} \times 3.25 cm$

= 37.7 $rad cm/s^{2}$

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 $rad cm/s^{2}$ .

6 0

3 years ago