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Ostrovityanka [42]
3 years ago
10

A box has a mass of 10.5 kg and has compressed a spring (with a spring constant of 71.7 N/m by 0.888m. The box is released from

rest on a frictionless surface
a. How fast is the box moving when it leaves the spring?

b. After the box has left the spring it comes to a surface with
Mk = 0.234 How fare does it slide on this surface before stopping?

c. Instead of sliding to a stop it only goes 0.432 m on the rough surface and then returns to a frictionless surface. How fast is it moving?
Physics
1 answer:
Luden [163]3 years ago
8 0

Answer:

a. To find how fast it moves when it leaves the spring we use

1/2kx²= 1/2 mv²,

So making speed v subject

v=x√(k/m),

v= 0.888√(71.7/10.5) = 2.32 m/s

B) how far it goes after released

We know that

Work done = change in kinetic energy

So

(1/2 mv²0)= F x D,

F= mu x N,

But N=mg , F=0.234 x 10.5x 9.81= 24.10

So

0.5x 10.5 x 2.32 x 2.32= D x 24.10,

So

D= 1.17 m

C. Now using the work energy theorm:

0.5(mv1²- mv2²) =F xD = mu. mg x D

V2=√(V1²- 2gmu) =

√(2.32x 2.32-2 x 0.432 x 9.81 x 0.234) =

V2=1.8m/s

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What is the voltage across each resistor?
kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30  = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55  = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15  = 5.4volts

Hence the voltage across the 15ohms resistor is 5.4volts

4 0
3 years ago
A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba
Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

\mu = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

m_1 = Mass of bean bag = 0.354 kg

m_2 = Mass of empty crate = 3.77 kg

v_1 = Speed of the bean bag

v_2 = Speed of the crate

Acceleration

a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g

a=--9.81\times 0.48=4.7088\ m/s^2

From equation of motion

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s

In this system the momentum is conserved

m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s

The speed of the bean bag is 31.42383 m/s

8 0
3 years ago
A _______ satellite records reflected wavelengths from Earth's surface.
Oksana_A [137]
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3 0
3 years ago
Read 2 more answers
john is using a pulley to lift the sail on his sailboat. the sail weighs 150N and he must lift it at 4.0m. how much work must be
gogolik [260]

The work done on the sail is 600 J

Explanation:

The work done to lift the sail is equal to the gain in gravitational potential energy of the sail, therefore is:

W=mg\Delta h

where

m is the mass of the sail

g is the acceleration of gravity

(mg) is the weight of the sail

\Delta h is the change in height of the sail

In this problem we have

mg = 150 N (weight)

\Delta h = 4.0 m

Substituting, we find the work done:

W=(150)(4.0)=600 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

5 0
3 years ago
At a constant pressure, 10 L of a gas at 546 K is cooled to 273 K
lisabon 2012 [21]

Answer:

5 L

Explanation:

Ideal gas law:

PV = nRT

If P, n, and R are constant, then:

n₁R/P₁ = n₂R/P₂

Using ideal gas law, we can rewrite this as:

V₁/T₁ = V₂/T₂

This is known as Charles' law.

Plugging in values:

10 L / 546 K = V / 273 K

V = 5 L

8 0
3 years ago
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