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2 years ago

how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance

Physics

1 answer:

Sergio039 [100]2 years ago

8 0

Answer:

W = F * s

Work done equals applied force * distance traveled

Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled

μ = coefficient of kinetic friction

F cos θ = force applied to motion of sled

s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

Note that F = μ M g if applied force is in the horizontal direction

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Answer: C. Chemical Weathering

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3 years ago

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What happens to the kinetic energy of a snowball as it rolls across the lawn and gains mass.

LiRa [457]

If the ground is level, then the snowball can never have
any more kinetic energy than it hand when it left your hand.

If more mass sticks to it as it makes its way across the lawn,
then it must slow down, so that its

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never exceeds the KE you gave it when you tossed it.

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4 years ago

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A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

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4 years ago

A block of mass M=10 kg is on a frictionless surface as shown in the photo attached. And it's attached to a wall by two springs

nordsb [41]

i. the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

ii. the angular velocity when the two springs are in parallel is 7.07 rad/s

i. the speed of the block of mass when the springs are connected in series is 11.2 A m/s

ii. the angular velocity when the two springs are in series is 11.2 rad/s

<h3>a. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in parallel?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is k' = k + k

= 2k

= 2 × 250 N/m

= 500 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k'A² = 1/2k'x² + 1/2Mv² where

k' = equivalent spring constant in parallel = 500 N/m,
A = maximum displacement of spring,
x = equilibrium position = 0 m,
M = mass of block = 10 kg and
v = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k'(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k'(A² - x²)/M]

v = √[500 N/m(A² - (0)²)/10]

v = √[50 N/m(A² - 0)]

v = [√50]A m/s

v = [5√2] A m/s

v = 7.07 A m/s

So, the speed of the block of mass when the springs are connected in parallel is 7.07 A m/s

<h3>ii. The angular velocity of mass when the springs are in parallel</h3>

Since velocity of spring v = ω√(A² - x²) where

ω = angular velocity of spring,
A = maximum displacement of spring and
x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 7.07 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 7.07 A m/s/√(A² - 0²)

ω = 7.07 A m/s/√(A² - 0)

ω = 7.07 A m/s/√A²

ω = 7.07 A m/s/A m

ω = 7.07 rad/s

So, the angular velocity when the two springs are in parallel is 7.07 rad/s

<h3>b. </h3><h3>i. How to calculate the velocity of the mass when the springs are connected in series?</h3>

Since k is the spring constant of both springs = 250 N/m. The equivalent spring constant in parallel is 1/k" = 1/k + 1/k

= 2/k

⇒ k" = k/2

k" = 250 N/m ÷ 2

= 125 N/m

Now since A is the maximum distance the block is pulled from its equilibrium position, the total energy of the block is E = 1/2kA

Also, 1/2k"A² = 1/2k"x² + 1/2Mv'² where

k" = equivalent spring constant in series = 125 N/m,
A = maximum displacement of spring,
x = equilibrium position = 0 m,
M = mass of block = 10 kg and
v' = speed of block at equilibrium position

Making v subject of the formula, we have

v = √[k"(A² - x²)/M]

Substituting the values of the variables into the equation, we have

v = √[k"(A² - x²)/M]

v = √[125 N/m(A² - (0)²)/10]

v = √[125 N/m(A² - 0)]

v = [√125]A m/s

v = [5√5] A m/s

v = 11.2 A m/s

So, the speed of the block of mass when the springs are connected in series is 11.2 A m/s

<h3>ii. The angular velocity of the mass when the springs are in series</h3>

Since velocity of spring v = ω√(A² - x²) where

ω = angular velocity of spring,
A = maximum displacement of spring and
x = equilbrium position of spring = 0 m

Making ω subject of the formula, we have

ω = v/√(A² - x²)

Since v = 11.2 A m/s

Substituting the values of the other variables into the equation, we have

ω = v/√(A² - x²)

ω = 11.2 A m/s/√(A² - 0²)

ω = 11.2 A m/s/√(A² - 0)

ω = 11.2 A m/s/√A²

ω = 11.2 A m/s/A m

ω = 11.2 rad/s

So, the angular velocity when the two springs are in series is 11.2 rad/s

Learn more about speed of block of mass here:

brainly.com/question/21521118

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2 years ago

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<h2>Hope it helps...</h2><h2>Have a great day : )</h2>

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3 years ago