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Viktor [21]
1 year ago
14

a model rocket flies horizontally off the edge of the cliff at a velocity of 50.0m/s.if the canyon below is 100.0m deep, how far

from the edge of the cliff does the model rocket land
Physics
1 answer:
Alona [7]1 year ago
8 0
For the winter, a duck flies 10.0 m/s due south against a gust of wind with a speed of 2.5 m/s. What is the resultant .... A model rocket flies horizontally off the edge of a cliff at a velocity of 50.0 m/s. If the canyon below is 100.0 m deep, how far from the edge of the cliff does the model rocket land? (a = −g = −9.81 m/s ). 45.
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The gravitational field of m1 is denoted by g1. Enter an expression for the gravitational field g1 at position la in terms of m1
Andrei [34K]

Answer:

The expression of gravitational field due to mass m_! at a distance l_a

Explanation:

We have given mass is m_1

Distance of the point where we have to find the gravitational field is l_a

Gravitational constant G

We have to find the gravitational filed

Gravitational field is given by g=\frac{Gm_1}{l_a^2}

This will be the expression of gravitational field due to mass m_! at a distance l_a

4 0
2 years ago
A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is
tiny-mole [99]

Answer:

F=mg(sin(\theta )-0.25 cos(\theta ))

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\

Using value of N from equation β in α we get value of force as

F=mg(sin(\theta )-\mu cos(\theta ))

Applying values we get

F=mg(sin(\theta )-0.25 cos(\theta ))

8 0
1 year ago
Read 2 more answers
Kenny and Candy decided to sit on a see-saw while visiting a local play park. Candy, of mass
pochemuha

Answer:

(i) 208 cm from the pivot

(ii) Move further from the pivot

Explanation:

(i) Sum of the moments about the pivot of the seesaw is zero.

∑τ = Iα

(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0

1250 Nm + 600 N x = 0

x = -2.08 m

Kenny should sit 208 cm on the other side of the pivot.

(ii) To increase the torque, Kenny should move away from the pivot.

4 0
2 years ago
The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f
tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

4 0
1 year ago
During a family trip to Laura's grandmother's house, the family cast traveled a distance of of 8 miles in 24 minutes. During the
Vanyuwa [196]

The correct answer is B

4 0
2 years ago
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