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Lena [83]
2 years ago
14

Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do

extra exercise to avoid gaining weight. He was doing jumps: he leaves the ground with a speed of 3.3m/s at every jump. Assuming that his body turns energy to mechanical work with a 25 % efficiency, how many jumps he will have to make g
Physics
1 answer:
Kaylis [27]2 years ago
4 0

Answer:

Explanation:

Calories to be burnt = 3500 - 2500 = 1000 Cals .

Efficiency of conversion to mechanical work  is 25 % .

Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.

4000 Cals = 4.2 x 4000 = 16800 J  .

Work done in one jump = kinetic energy while jumping

= 1/2 m v²

= .5 x 70 x 3.3²

= 381.15 J .

Number of jumps required = 16800 / 381.15

= 44 .

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Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
kodGreya [7K]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

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2 years ago
Volcanoes are often formed at plate boundaries. This is a convergent plate boundary. From the choices listed, pick the correct d
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Your answer should be a
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1. A car is 140 kg, and drove East 13.5 m/s. Car B is 157 kg and drove West at 10.9 m/s.
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The velocity would switch on the cars
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2 years ago
A person walks 25 m west and then 45 m at the angle of 60 degrees north of east what is the magnitude of the total displacement?
expeople1 [14]


To solve this question, we need to use the component method and split our displacements into their x and y vectors. We will assign north and east as the positive directions.

The first movement of 25m west is already split. x = -25m, y = 0m.

The second movement of 45m [E60N] needs to be split using trig.
x = 45cos60 = 22.5m
y = 45sin60 = 39.0m

Then, we add the two x and two y displacements to get the total displacement in each direction.

x = -25m + 22.5m = -2.5m
y = 0m + 39.0m

We can use Pythagorean theorem to find the total displacement.
d² = x² + y²
d = √(-2.5² + 39²)
d = 39.08m

And then we can use tan to find the angle.
inversetan(y/x) = angle
inversetan(39/2.5) = 86.3

Therefore, the total displacement is 39.08m [W86.3N]

8 0
2 years ago
A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
Elan Coil [88]

Answer:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.  </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.  </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we use our previous value for the work:

W_{ff}=-F_f*(hy/sin\theta)   Solving for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

Vf^2=Vo^2+2*a*d

Solving for a:

a=-3.844m/s^2

Now, by dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
2 years ago
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