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Westkost [7]
3 years ago
11

Please, pretty please help me!

Physics
2 answers:
Andrews [41]3 years ago
8 0

Answer:

umm

Explanation:

1223565 578633 =334675

Harrizon [31]3 years ago
7 0
Hhsjjsjdbdvbr fineee
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"The White Shark" allows riders to start from rest on a tube and then slide down a 44 meter slide. It takes the rider 6.2 second
Leni [432]
<span>Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. Calculations of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:

d = v0t + at^2/2

We cancel the term with v0 since it is initially at rest,

d = at^2/2
44 = a(6.2)^2/2
a = 2.3 m/s^2



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6 0
3 years ago
What are the opposite ends of a magnet called? A. north and south terminals B. north and south poles C. magnetic fields D. magne
Bezzdna [24]

I think the answer is B.

Hope this helps.

6 0
3 years ago
Read 2 more answers
A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra
bija089 [108]

Explanation:

Terminal velocity is given by:

v_t=\sqrt{\frac{2mg}{\rho C_dA}}

Here, m is the mass of the falling object, g is the gravitational acceleration,  C_d is the drag coefficient, \rho is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is 1.28\frac{kg}{m^3}.

v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}

7 0
3 years ago
What is the connection between change in momentum and impulse
Katena32 [7]
The change in momentum of an object equals the impulse applied to it
6 0
3 years ago
A ball is held at rest at the top of a hill. The ball is then released and starts rolling down the hill. At the bottom it reache
Alex17521 [72]

Answer:

Gravitational Potential Energy

Explanation:  

a ball is held rest at the top of hill  

gravitational potential energy will store due to its height  

it.   and  body will start move downward and its potential energy will convert into kinetic energy due to motion of body

at the ground level it will stop and potential energy will became zero and kinetic energy get convert into internal energy due to collisions

3 0
3 years ago
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