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3 years ago

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Physics

1 answer:

mihalych1998 [28]3 years ago

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The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

7 0

3 years ago

The value of acceleration due to gravity (g) on Pluto is about . 0.61 meters/second2. How much will an object that weighs 250 ne

Annette [7]

First, determine the mass of the object by dividing its weight on Earth by 9.8 m/s² as shown below,
m = 250 N / 9.8 m/s² = 25.51 kg
Then, multiply the obtained mass by the acceleration due to gravity (g) on Pluto.
W (in Pluto) = (25.51 kg) x (0.61 m/s²) = 15.56 N
Therefore, the object will only weigh 15.56 N.

8 0

3 years ago

A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.

zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = $7.50\times 10^{- 3}\ H$

Current in the coil, i = $1680cos[\frac{\pi t}{0.0250}]$ A

where

$i_{max} = 1680\ mA = 1.680\ A$

Now,

(a) To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

$e = \frac{Ldi}{dt}$

$e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]$

$e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]$

For maximum emf, $sin\theta$ should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

$|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V$

(b) To calculate the maximum average flux,we know that:

$\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb$

$e = e_{o}sin{\pi t}{0.0250}$

$e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V$

7 0

3 years ago

How does the law of conservation of energy apply to machines?

trasher [3.6K]

The answer is letter C

7 0

3 years ago

Read 2 more answers

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago