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mamaluj [8]
2 years ago
5

How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav

itational acceleration on the Moon is about 1/6 of g on Earth)?
Physics
1 answer:
JulsSmile [24]2 years ago
8 0

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

v_f^2 - v_i^2 = 2 aH

now from above equation we will have

0 - v^2 = 2(-g)H

now we have

H = \frac{v^2}{2g}

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

0 - v^2 = 2(-g/6) H

now we have

H = \frac{6v^2}{2g}

so maximum height on moon is 6 times more than the maximum height on Earth

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bazaltina [42]

The energy released from glucose during this reaction is used and also wasted. Some of the energy is used as work while some amounts are used in other processes or stored for transfer to other organisms. Also, some of the energy is wasted in the form of heat.

6 0
1 year ago
A car has an initial velocity of 11.2 m /sec. the car accelerates at 10.0 m /s2 for 8.0 seconds. what is the velocity of the car
Vinil7 [7]
Initial velocity(u) = 11.2 m/s.
Final velocity(v) = ?
acceleration(a) = 10.2 m/s²

Using kinematic equation v = u + at

v = 11.2 + 10 x 8 = 11.2 + 80 = 91.2 m/s.

Therefore final velocity is 91.2 m/s.
7 0
3 years ago
Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0°
Tpy6a [65]

Answer:

Distance between slits, d=2.89\times 10^{-7}\ m  

Explanation:

It is given that,

Wavelength, \lambda=410\ nm=410\times 10^{-9}\ m

Angle, \theta=45

We need to find the distance between two slits that produces first minimum. The equation for the destructive interference is given by :

d\ sin\theta=(n+\dfrac{1}{2})\lambda

For first minimum, n = 0

So, d\ sin\theta=(\dfrac{1}{2})\lambda

d is the distance between slits

So, d=\dfrac{1/2\lambda}{sin\theta}

d=\dfrac{1/2\times 410\times 10^{-9}}{sin(45)}

d=2.89\times 10^{-7}\ m

So, the distance between two slits is 2.89\times 10^{-7}\ m. Hence, this is the required solution.

6 0
3 years ago
A lever is used to lift a boulder. The fulcrum is placed 1.60 m away from the end at which you exert a downward force, producing
KengaRu [80]
Ion even know maybe a,b,c,
5 0
2 years ago
Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

x_f = Final length of the spring .

= \sqrt{1.25^2+1.8^2}  -0.60

= 1.591 m

x_i= The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

=  1.80 m.

m= The mass of the collar.

=3.55 kg

v_c = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Or, \frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2}  = 3.55\times 9.81\times 1.80

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

6 0
2 years ago
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