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3 years ago

How much does a 2 kg apple weigh? Enter in your numerical answer.

Physics

2 answers:

Monica [59]3 years ago

8 0

umm, something tells me it will weight 2 kg

murzikaleks [220]3 years ago

6 0

Answer:

2,000 grams, 0.002 metric tons, not sure what you are asking for, but hopefully this helps

You might be interested in

A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.

Natali5045456 [20]

Answer:

19.62 ms

Explanation:

t = Time taken = 2 s

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive)

Equation of motion

$v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ m/s$

The speed of the pebble when it hit the water is 19.62 ms

3 0

3 years ago

Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. What is the min

Paha777 [63]

Answer: The correct answer is option (A).

Explanation:

Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s

$P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s$

Then the maximum speed of the another car in order to not to break the eggs will be same as first car:

$P_1=P_2$

$0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2$

$v_2=1 m/s$

Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.

4 0

4 years ago

In which condition the final velocity becomes zero?

alexgriva [62]

Answer:

Terminal velocity is achieved, therefore, when the speed of a moving object is no longer increasing or decreasing; the object's acceleration (or deceleration) is zero.

Explanation:

hope it helps you

7 0

3 years ago

The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar m

lions [1.4K]

Answer:

Explanation:

a. Landing height is

H=1.3m

Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft

u=1.3m/s

Velocity of lander at impact, i.e final velocity is needed

v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

v²=11.8924

v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m

8 0

3 years ago

What is 3.75 x 10^-7?

pogonyaev

Answer:

Explanation:

3.75 * 10^-7

=3.75 * 1/10^7

=3.75/10000000

=3/800000000

any base which has it's power negative do it's reciprocal then the power will be positive.

8 0

3 years ago