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Tju [1.3M]
3 years ago
14

100 J of work in 10 seconds

Physics
1 answer:
HACTEHA [7]3 years ago
7 0
The average power is 10 watts. What's the question ?
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A block of mass 4 kg slides down an inclined plane inclined at an angle of 30o with the horizontal. Find the acceleration of the
Nataly [62]

Answer:

a) a = 4.9 m/s²

b) a = 1.5 m/s²

Explanation:

no friction

F = ma

gsinθ = ma

a = gsinθ

a = 9.8sin30

a = 4.9 m/s²

friction

gsinθ - μmgcosθ = ma

a = g(sinθ - μcosθ)

a = 9.8(sin30 - 0.4cos30)

a = 1.5051...

8 0
3 years ago
What is weight in Newton’s, of a 50.-kg person on earth
Svetlanka [38]

Answer:

It is about 490 Newtons. 490.3325 to be exact

Explanation:

Pls mark as the Brainliest answer. Thank You very much, enjoy your day

8 0
3 years ago
List examples of how the Bill of Rights protects you:<br> .<br> .<br> .<br> .<br> .
Naya [18.7K]

guaranteeing freedom of speech, press, assembly, and exercise of religion

4 0
2 years ago
The mass of a certain neutron star is 2.5x10^30kg and the radius 7000m. what is the force of gravity on a 1kg object of the surf
Makovka662 [10]

Answer:

3.42N

Explanation:

*not too sure bc i left my physics notes at school so it might not be 100% accurate :p*

Use the equation: F = (GMm)/(r^2)

F = force of gravity

G = gravitational constant (6.7x10^-11)

M = mass1 (2.5x10^30kg)

m = mass2 (1kg)

r = radius (7000m)

Plug it in: F = ((6.7x10^-11)(2.5x10^30)(1)) / (7000^2)

F = (1.675x10^20) / (4.9x10^7)

F = 3.4183673x10^12

F = 3.42N

8 0
3 years ago
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
3 years ago
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