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V125BC [204]
3 years ago
15

Your roommate drops your wallet down to you from the third-floor window of your apartment, which is 11.5 m from the ground. What

is the velocity of the wallet when you catch it? (assume you catch the wallet right before it hits the ground)
Physics
1 answer:
Aneli [31]3 years ago
6 0

Answer:

15 m/s

Explanation:

Using the law of conservation of energy, potential energy equals kinetic energy hence

mgh=0.5mv^{2}

Therefore

v=\sqrt{2gh}

where g is the acceleration due to gravity, m is the mass of the object, h is the height and v is the speed of the wallet

Taking g as 9.81 then

v=\sqrt{2\times 9.8\times 11.5}=15.02098532  m/s\approx 15 m/s

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Is cell using oxygen to break down sugar a physical or chemical change
Viefleur [7K]
Breaking down sugar (glucose) is a chemical change. Sugar is a compound that can be broken down.
5 0
3 years ago
Read 2 more answers
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun
Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{100.0}{300}

t=0.33\ sec

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

s=0+\dfrac{1}{2}\times9.8\times(0.33)^2

s=0.53\ m

Hence, The distance is 0.53 m.

5 0
3 years ago
A 0.290 kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato
Sergeu [11.5K]

Answer:

A) The speed of the potato at the lowest point of its motion is 7.004 m/s

B) The tension on the string at this point is 8.5347 N

Explanation:

Here we have that the height from which the potato is allowed to swing  is 2.5 m

Therefore we have ω₂² = ω₁² + 2α(θ₂ - θ₁)

Where:

ω₂ = Final angular velocity

ω₁ = Initial angular velocity = 0 rad/s

α = Angular acceleration

θ₂ = Final angle position

θ₁ = Initial angle position

However, we have potential energy of the potato

= Mass m×Gravity g× Height h

= 0.29×9.81×2.5 = 7.1125 J

At he bottom of the swing, the potential energy will convert to kinetic energy as follows

K.E. = P.E. = 7.1125 J

1/2·m·v² = 7.1125 J

Therefore,

v² = 7.1125 J/(1/2×m) = 7.1125 J/(1/2×0.290) = 49.05

∴ v = √49.05 = 7.004 m/s

B) Here we have the tension given by

Tension T in the string = weight of potato + Radial force of motion

Weight of potato = mass of potato × gravity

Radial force of motion of potato = mass of potato × α,

where α = Angular acceleration = v²/r and r = length of the string

∴ Tension T in the string = m×g + m×v²/r = 0.290×(9.81 + 7.004²/2.5)

T = 8.5347 N

4 0
3 years ago
Read 2 more answers
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t
coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

    Y axis

     N- W = 0

     N = W = mg

  X axis

     -Fr = m a

     -μ N = m a

     -μ mg = ma

     a = μ g

     a  = - 0.32 9.8

     a =  - 3.14 m/s²

We calculate the distance using the kinematics equations

    Vf² = Vo² + 2 a x

     x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

 Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

     x = ( 0 - 15²) / 2 (-3.14)

     x=  35.8 m

The shortest braking distance is  35.8 m

7 0
3 years ago
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp
Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J

So, the new work is more than 130 J.

6 0
2 years ago
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