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pishuonlain [190]
3 years ago
14

A monkey has a bit of a heavy for on the gas pedal. As soon as the light turns green the monkey pushes the gas pedal to the floo

r and accelerates from 0 mph to 60 mph (0 m/s - 28.82 m/s) in 4.2 seconds, calculate the cars acceleration.
Physics
1 answer:
Andrei [34K]3 years ago
3 0

Answer:

s=6.86m/s^2

Explanation:

Hello,

In this case, considering that the acceleration is computed as follows:

a=\frac{v_{final}-v_{initial}}{t}

Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:

a=\frac{28.82m/s-0m/s}{4.2s}\\ \\s=6.86m/s^2

Regards.

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While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into a room an
Bogdan [553]

Answer:

the time needed for her to close the door is 1.36 s.

Explanation:

given information:

Force, F = 220 N

width, r = 1.40 m

weight, W = 790 N

height, h = 3.00 m

angle, θ = 90° = π/2

to find the times needed to close the door we can use the following equation

θ = ω₀t + 1/2 αt²

where

θ = angle

ω = angular velocity

α = angular acceleration

t = time

in this case, the angular velocity is zero. thus,

θ = 1/2 αt²

now, we can find the angular speed by using the torque formula

τ = I α

where

τ = torque

I = Inertia

we know that

τ = F r

and

I = 1/3 mr²

so,

τ = I α

F r = 1/3 mr² α

α = 3 F/mr

  = 3 F/(w/g)r

  = 3 (220)/(790/9.8) 1.4

  = 5.85 rad/s²

θ = 1/2 αt²

π/2 = 1/2 5.85 t²

t = 1.36 s

5 0
3 years ago
If the spring constant is doubled, what value does the period have for a mass on a spring?
soldi70 [24.7K]

if spring constant is doubled, the mass on spring will be doubled as well. according to this formula, F=ke

k stands for spring constant and e stands for the length extended

4 0
3 years ago
Read 2 more answers
It is estimated that 1kg of body fat will provide 3.8 * 10^7 J of energy. A 67kg mountain climber decides to climb a mountain 35
xz_007 [3.2K]
During a climb UP the mountain, gravity does NO work on the climber.
Actually, it's more correct to say that gravity does NEGATIVE work
on him.  The climber has to DO the positive work to haul himself up.
  
                     Work = (mass) x (gravity) x (height) .

For the guy in this problem:

                     Work = (67 kg) x (9.8 m/s²) x (3,500 meters)

                             =  2,298,100 joules.

If he eats no candy bars on the way, and completely depends on
his stored body fat for the energy, then he'll burn off

                       (2,298,100 joules) / (3.8 x 10⁷ joules/kg)

                   =          0.06 kg of fat.

That's only about 2.1 ounces.  We KNOW he'll lose more weight than that,
climbing 11,000 feet.  That's because climbing is pretty inefficient. 
In addition to the potential energy you have to give your body weight,
you also have to expend energy breathing, digesting, metabolizing,
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4 0
3 years ago
Two charges are sitting 1.5 m apart with a force of 3 N between them. They are now moved farther apart to 2.25 m and one of the
ANTONII [103]

Answer: 2.37N

Explanation:

According to coulombs law which states that the force of attraction (F) between two charges (q1 and q2) is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them. Mathematically,

F = kq1q2/r²

For the first two charges that are sitting 1.5 m apart with a force of 3 N between them, we have

3 = kq1q2/1.5²

3 = kq1q2/2.25

Kq1q2= 6.75... (1)

If the charges are now moved farther apart 2.25 m and one of the charges is increased by a factor of 4. The formula becomes

F2 = k(4q1)q2/2.25² (q1 has been increased by factor of 4)

k(4q1)q2 = 5.06F2 ... (2)

Dividing 2 by 1 we have

k(4q1)q2/kq1q2 = 5.06F2/3

4 = 5.06F2/3

5.06F2 = 12

F2= 12/5.06

F2 = 2.37N

Therefore the magnitude of the new force between the two charges is 2.37N

5 0
3 years ago
Consider the following setup with three identical springs, a ruler for length measurements and three known masses and three unkn
svetlana [45]

Answer:

To find the value of the unknown weight, we previously placed the 3 known weights and made a graph of the force against displacement

When hanging the weight is known, we measure the displacement and from the graph we can find the value of the hanging masses

We can also use the equation and multiply the constant K by the displacement and this is the applied weight.

Explanation:

For this problem we will use the translational equilibrium relation

        F –W = 0

        F = W

        W = mg

The spring elastic force is

        F = - k x

We substitute

        k x = m g

Where we see that the force of the spring is equal to the weight of the body.

To find the value of the unknown weight, we previously placed the 3 known weights and made a graph of the force against displacement

When hanging the weight is known, we measure the displacement and from the graph we can find the value of the hanging masses

We can also use the equation and multiply the constant K by the displacement and this is the applied weight.

4 0
3 years ago
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