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nikitadnepr [17]
3 years ago
6

What what is the direction of an object’s acceleration if it is being pulled upward by a rope at a constant speed?

Physics
1 answer:
Arada [10]3 years ago
7 0

Answer:

the direction is towards upward.i think.

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A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k
9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg  : mass of the  skier

θ =30°  :angle θ of the ski slope  with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W=  92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N=  0.10*780.8 N  

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 =  (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0
3 years ago
A student wants to use a ramp to move boxes into a truck bed that is 3 m high. He has a choice of 2 different ramps. The length
Morgarella [4.7K]

Answer:

The correct option is;

B. 8 m, because he has to apply less force over a greater distance

Explanation:

In the given question, in order for the student to lift the boxes onto the tuck with less amount of force, he applies the principle of Mechanical Advantage

The mechanical advantage is given by the measure by which a force is amplified through the use of a tool

Given that the work done = The force × The distance, we have

F₁ × d₁ = F₂ × d₂, which gives;

d₁/d₂ = F₂/F₁

Where;

F₁ = The input force

F₂ = The output force

d₁ = The input distance

d₂ = The output distance

The Mechanical advantage, MA = d₁/d₂ = F₂/F₁

Therefore, when the input distance is increased the input force will be reduced for a given output force

6 0
3 years ago
Black holes are:
agasfer [191]
Answer to the question provided is A
4 0
4 years ago
Read 2 more answers
PLEASE HELP ME ASAP. IT'S VERY IMPORTANT
hodyreva [135]

Answer:

1) a.  52.41 m/s

b. The skier will be going 15.35 m/s slower

2)  103.68 m

3) 35,127 J

4) a.  88.825 kJ

(b) 16.36 %

5) 3,071.12 J

Explanation:

1) a. The given height of the hill, h = 140.0 m

The mass of the skier at the top of the hill, m = 85.0 kg

The acceleration due to gravity, g = 9.81 m/s²

The initial potential energy, P.E of the skier = m×g×h = 85.0×140.0×9.81 = 116739 J

From the principle of conservation of energy, we have;

The potential energy, P.E. lost = The gain in kinetic energy, K.E.

m×g×h = 1/2×m×v²

116739 J = 1/2×85.0×v²

v² = 116739/(1/2*85.0)= 2746.8 m²/s²

v = √(2746.8 m²/s²) = 52.41 m/s

b. From 70 m up, we have;

The initial potential energy, P.E., of the skier is now = 85.0×70×9.81 = 58,369.5 J

The potential energy, P.E. lost = The gain in kinetic energy, K.E.

58,369.5 J = 1/2×85.0×v²

v² = 58,369.5/(1/2*85.0) = 1373.4 m²/s²

v = 37.06 m/s

The skier will be going 52.41 - 37.06 = 15.35 m/s slower

The skier will be going 15.35 m/s slower

2) From the principle of conservation of energy, the amount of work done (energy used) = The (potential) energy gained by the load

The amount of work done by the electric hoist = 356,000 J

The mass of the load = 350.0 kg

The height to which the load is raised = h

The potential energy gained by the load = m×g×h = 350.0×9.81×h

356,000 J = 350.0×9.81×h

h = 356,000/(350.0*9.81) = 103.68 m

The height to which the load is lifted= 103.68 m

3) The initial potential energy of the roller coaster cart = 600*35.0*9.81 = 206010 J

The final potential energy = 600*28.0*9.81= 164808 J

The velocity at point 3  = 4.5 m/s

The kinetic energy at point 3 = 1/2*600*4.5^2 = 6075 J

The total energy at point 3 = 164808 + 6075 = 170,883 J

The energy loss = The initial potential energy at point 1 - Total energy at point 3

The energy loss = 206010 - 170,883 = 35,127 J

The heat energy due to friction that must have been produced between points 1 and 3 = 35,127 J

4) a. The heat energy absorbed = mass × specific heat capacity for water, C_{water} × Temperature change

The mass of the water = 2.5×10² g = 0.25 kg

C_{water} = 4,180 J/(kg·°C)

Initial temperature = 10.0°C

Final temperature = 95°C

The temperature change = 95.0°C - 10.0°C = 85.0°C

The heat energy absorbed = 0.25*4,180* 85 = 88,825 J = 88.825 kJ

(b) The percentage efficiency = (Heat absorbed/(Heat supplied)) × 100

The heat supplied = 543 kJ

The efficiency = (88.825/543) × 100  = 16.36 %

5) The mass of the box = 115 kg

Force acting on the rope = 255 N

The angle of inclination of the force to the horizontal = 24.5°

The distance the box is displaced = 15.0 m to the right

The work done = Force applied × distance moved in the direction of the force

The work done = Force applied × distance moved in the direction of the force

Given that the load moves a distance 15.0 m to the right,we have;

The component of the force acting in the direction of the movement of the load (to the right) is 225 × cos(24.5°) =  204.74 N

The work done = 204.7*15 = 3071.12 J

The amount of work done  = 3,071.12 J

6 0
3 years ago
Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp
maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = \dfrac{N}{C}m

c = Speed of light = m/s

Momentum is given by

\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s

The unit of MeV/c in SI fundamental units is kg m/s

5 0
3 years ago
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