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2 years ago

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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

1 answer:

Alex Ar [27]2 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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1. A bicycle initially moving with a velocity

ki77a [65]

Answer:

$\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}$

Explanation:

We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.

$v_f= v_i+ at$

In this formula, $v_f$ is the final velocity, $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.

$\bullet \ v_i = 5.0 \ m/s \\\bullet \ a= 2\ m/s^2\\\bullet \ t= 5 \ s$

Substitute the values into the formula.

$v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)$

Solve inside the parentheses.

$\frac {2 \ m}{s^2}* 5 \ s = \frac{ 2 \ m}{s} * 5 = \frac{ 10 \ m}{s} = 10 \ m/s$

$v_f= 5.0 \ m/s + (10 \ m/s)$

Add.

$v_f= 15 \ m/s$

The units can also be written as:

$v_f= 15 \ m*s^{-1}$

The bicycle's final velocity is 15 meters per second.

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2 years ago

I dont get why its c i dont under the graph for the vertical component in option c

jek_recluse [69]

The vertical component grows at an increasing rate because
gravity is ACCELERATING the object vertically. So its speed
keeps growing. Speed is the slope of the displacement graph.

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3 years ago

Neglecting air resistance, when does a struck baseball accelerate downward at a rate of 9.8 m/s2?

olga2289 [7]

A baseball is accelerated downward at 9.8 m/s^2 when the only vertical force acting upon the baseball is gravity.

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3 years ago

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3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What

WITCHER [35]

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

$F_{net}=ma$

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction $F_f$ , opposing the motion of the box, so to the left

So we can write the net force as

$F_{net}=F-F_f$

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

$a=0$

Therefore

$F_{net}=0$

WHich means:

$F-F_f=0$

And therefore,

$F_f=F=25 N$

which means that the force of friction is also 25 N.

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2 years ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

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3 years ago