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Bond [772]
3 years ago
15

This is a textbook question, work too please!

Physics
1 answer:
lara [203]3 years ago
4 0
The force is not pulling in the direction that the box is moving.
Only the force in the horizontal direction does work.
The force in the vertical direction doesn't do any work, because
       the box doesn't move through any vertical distance.

The horizontal force is  200(cos 45°)  =  142 newtons

The vertical force is  200(sin 45°)  =  142 newtons

If you pull the box 5 meters, then the amount of work done is

           (force) x (distance) = (142 N) x (5 m) = 710 joules.

Power  =  (work done) / (time to do the work)

           =  (710 joules)  /  (5 seconds)  =  142 watts .
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An electric field from a charge has a magnitude of 1.5 × 104 N/C at a certain location that points inward. If another charge wit
zysi [14]

Answer:

-0.045 N, they will attract each other

Explanation:

The strength of the electrostatic force exerted on a charge is given by

F=qE

where

q is the magnitude of the charge

E is the electric field magnitude

In this problem,

q=3.0\cdot 10^{-6}C

E=-1.5\cdot 10^4 N/C (negative because inward)

So the strength of the electrostatic force is

F=(-3.0\cdot 10^{-6}C)(1.5\cdot 10^4 N/C)=-0.045 N

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.

6 0
3 years ago
Read 2 more answers
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
Dalton proposed the first atomic model. Which of these statements accurately reflected his thinking at that time?
Sunny_sXe [5.5K]
B. Atoms are solid balls
3 0
3 years ago
Read 2 more answers
A grindstone of mass 12 kg and radius 0.3 m is initially rotating freely at 48 rad/sec. An axe is brought into contact with the
lesya [120]

Answer:

I = 0.54\,kg\cdot m^{2}

Explanation:

1) The moment of inertia of the grindstone is:

I = \frac{1}{2}\cdot m \cdot r^{2}

I = \frac{1}{2}\cdot (12\,kg)\cdot (0.3\,m)^{2}

I = 0.54\,kg\cdot m^{2}

4 0
3 years ago
Read 2 more answers
A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600
3241004551 [841]

Answer:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

The initial total mass of the rocket is:

m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

7 0
3 years ago
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