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DaniilM [7]
3 years ago
14

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

upon landing to reduce the force of the impact. A 73.0 kg man just before contact with the ground has a speed of 6.46 m/s. In a stiff-legged landing he comes to a halt in 2.07 ms. Calculate the average net force that acts on him during this time
Physics
1 answer:
Alex Ar [27]3 years ago
7 0

Answer:

Explanation:

The man comes to halt due to reaction force acting on him in opposite direction . If R be the reaction force

impulse by net  force = change in momentum

Net force = R - mg , mg is weight of the man .

( R-mg ) x 2. 07 x 10⁻³ = 73 x 6.46 - 0

R - mg = 227.81 x 10³

Average net force = 227.81 x 10³ N .

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3 years ago
A soccer player practices kicking the ball into the goal from halfway down the soccer field. The time it takes for the ball to g
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This is correct, I just did the test. Yes, displacement is 45 meters, elapsed time is three seconds, and the direction is toward the goal.

4 0
3 years ago
Read 2 more answers
State coulombs law in word​
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<h2><em>state coulombs law in word</em></h2>

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3 years ago
A 40 cm wire with a radius of 3 cm is oriented along the y axis and carries a current of 2 A. What is the magnitude of the magne
ZanzabumX [31]

Answer:

a) B = 1.99 x 10⁻⁴ Tesla

b) B = 0.88 x 10⁻⁴ Tesla

Explanation:

According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:

B = μ₀ I L/4πr²

where,

μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹

I = current = 2 A

L = Length of wire = 40 cm = 0.4 m

a)

r = radius of magnetic field = 2 cm = 0.02 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²

<u>B = 1.99 x 10⁻⁴ Tesla</u>

<u></u>

b)

r = radius of magnetic field = 3 cm = 0.03 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²

<u>B = 0.88 x 10⁻⁴ Tesla</u>

7 0
3 years ago
A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

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v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

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where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
2 years ago
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