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kap26 [50]
3 years ago
11

Please HELP! Many points! How much work does a weightlifter do if he lifts a 400 Newton barbell 3 meters high- 15 times???

Physics
1 answer:
77julia77 [94]3 years ago
4 0
Work = (force) x (distance)

1 time:   (400 newtons) x (3 meters)  =  1,200 joules

15 times:  (1,200 joules) x (15)  =  18,000 joules
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An air track car with a mass of 0.75 kg and a velocity of 8.5 m/s to the right collides elastically with a 0.65kg car moving to
Sunny_sXe [5.5K]
We can do this with the conservation of momentum. The fact it is elastic means no KE is lost so we don't have to worry about the loss due to sound energy etc.

Firstly, let's calculate the momentum of both objects using p=mv:

Object 1:
p = 0.75 x 8.5 = 6.375 kgm/s

Object 2 (we will make this one negative as it is travelling in the opposite direction):
p = 0.65 x -(7.2) = -4.68 kgm/s

Based on this we know that the momentum is going to be in the direction of object one, and will be 6.375-4.68=1.695 kgm/s

Substituting this into p=mv again:

1.695 = (0.75+0.65) x v
Note I assume here the objects stick together, it doesn't specify - it should!

1.695 = 1.4v
v=1.695/1.4 = 1.2 m/s to the right (to 2sf)
8 0
3 years ago
Determine the value of the force exerted by the surface (normal force) on a
Advocard [28]

Answer:

53.5 N

Explanation:

Vertical component of the F force   50 sin30    = 25 N  upward

   force of gravity = m g = 8 * 9.81 =78.5 N Downward

NET downward force by block on table = net upward force exerted by table =  78.5 -25 =53.5 N

8 0
2 years ago
Q
GalinKa [24]

Answer:

B

Explanation:

It would be diffrent if on a downward slope but assuming your going straight it would be the smallest student.

3 0
3 years ago
An astronaut circling the earth at an altitude of 400 km is horrified to discover that a cloud of space debris is moving in the
elena-14-01-66 [18.8K]

One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.

From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.

By equilibrium the centrifugal force and the gravitational force are equal therefore

F_c = F_g

\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}

Where

m = mass spacecraft

v = velocity

G = Gravitational Universal Constant

M = Mass of earth

r \rightarrow R+h \Rightarrow Radius of earth and orbit

Re-arrange to find the velocity

\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}

\frac{v^2_{orbit}}{r} = \frac{GM}{r^2}

v^2_{orbit}=\frac{GM}{r}

v_{orbit} = \sqrt{\frac{GM}{r}}

v_{orbit} = \sqrt{\frac{GM}{R+h}}

Replacing with our values we have

v_{orbit} = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{6.37*10^6+0.4*10^6}}

v_{orbit} = 7676m/s

From the cinematic equations of motion we have to

t = \frac{d}{2v_{orbit}} \rightarrow Remember that the speed is double for the counter-direction of the trajectories.

Replacing

t = \frac{29000m}{7676m/s}

t = 3.778s

Therefore the time required is 3.778s

4 0
3 years ago
What happens to a visible light wave when you increase the frequency
Anastasy [175]
When frequency increases, the wavelength halves.
6 0
3 years ago
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