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elena-14-01-66 [18.8K]

3 years ago

5

A soccer ball is kicked horizontally and rolled off of a 30 meter high cliff. The soccer ball was kicked with a velocity of 17m/

s. How long did it take before the ball hit the ground? How far from the cliff did the ball land? (Make sure to include units.)
VERTICLE
Acceleration =
Distance =
Initial velocity=
Time=
HORIZONTAL
Distance=
Velocity=
Time=
Show your work when solving for time:

Show your work when solving for horizontal distance:

1 answer:

Montano1993 [528]3 years ago

8 0

Answer:

i am doing this rn

Explanation:

COMMENTS

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Holding onto a tow rope moving parallel to a frictionless ski slope, a 61.8 kg skier is pulled up the slope, which is at an angl

nata0808 [166]

Answer:

a) Frope= 71.7 N

b) Frope=6.7 N

Explanation:

In the figure the skier is simulated as an object, "a box".

a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:

∑F=0

Frope=w*sen6.8°

Frope=71.71N

Take into account that w is the weight that is calculated as mass per gravitiy constant:

w=m*g

$w=61.8Kg*9.8\frac{m}{s^{2} }$

$w=605.64N$

b) In this case the system has an acceleration of 0.109m/s2. Then, we apply Newton's second law of motion:

F=m*a

F=61.8Kg*0.109m/s2

Frope=6.73N

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3 years ago

2. If you are 5'10" tall, that is, 5 feet 10 inches, what is your height in meters? (2 54 cm = 1.00 in)

likoan [24]

Answer:D 1.7

Explanation:

Just trust me

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3 years ago

Read 2 more answers

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

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3 years ago

You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after

kotykmax [81]

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = $\frac{W}{F}$

d = $\frac{3}{9}$

d = 0.33 m

The distance d the block ice moved is 33 cm.

7 0

3 years ago

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position

antiseptic1488 [7]

Answer:

The value of the distance is $\bf{14.52~cm}$ .

Explanation:

The velocity of a particle(v) executing SHM is

$v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)$

where, $\omega$ is the angular frequency, $A$ is the amplitude of the oscillation and $x$ is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., $x = 0$ .

The maximum velocity( $\bf{v_{m}}$ ) is

$v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Divide equation (1) by equation(2).

$\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Given, $v = 0.25 v_{m}$ and $A = 15~cm$ . Substitute these values in equation (3).

$&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm$

6 0

3 years ago