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3 years ago

13

PLEASE HELP PHYSICS

1 answer:

liberstina [14]3 years ago

7 0

Answer:

v = 11.7 m / s

Explanation:

For this exercise we can use the conservation of energy

starting point. Where you drop the stone

Em₀ = K + U = ½ m v² + m g h₁

final point. Point where the stone arrives

Em_f = U = mgh₂

We are assuming that there is no friction, so the energy is conserved

Em₀ = Em_f

½ m v² + m g h₁ = m g h₂

v² = 2g (h₂-h₁)

let's calculate

v² = 2 9.8 (17 - 10)

v = √137.2

v = 11.7 m / s

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What is the primary cause of diffusion?

Lostsunrise [7]

<span><u>Answer</u>
c). random internal motion of atoms and molecules.

<u>Explanation </u>
Diffusion is the movement of particles from a region of high concentration to region of low concentration. The rate of diffusion can be increased by increasing amount of temperature but this is not its primary cause. Diffusion is primarily caused by movement of molecules or atoms in a substances. Form the choices given, the correct answer is c. random internal motion of atoms and molecules.
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3 0

3 years ago

Read 2 more answers

what is the total distance traveled by a bike rider who rides for three hours at 40 km/hr and then two more hours at 50 km/hr?

marta [7]

Answer:

220 km

Explanation:

3 hours at 40 km/hr

40 km = 1 hour

40 km times 3 = 120 km

2 hours at 50 km/hr

50 km = 1 hour

50 times 2 = 100 km

Total distance

120 km + 100 km = 220 km

4 0

3 years ago

A pump is used to lift 100 KG of water from a wel 60 m deep,in 20 S If force of gravity on 1 KG is 10 N,find

Harlamova29_29 [7]

Explanation:

Given,

m = 100 kg
g = 10 N/kg¹
h = 60 m
t = 20 s

To Find:

a) Work done by the pump

b) Potential energy stored in the water

c)Power spent by the pump

d)Power rating of the pump.

Solution:

a) Work done by the pump

We know that,

$\rm \: Work \: done = Force * Distance \: moved$

f = 100 kg * 10N/kg
d = 60 m

$\rm \: Work\; Done =(100 \: kg \times \cfrac{10N}{kg} ) \times 60 \: m$

$\rm \: Work\; Done =1000 \times 60 \: joule$

$\boxed{\rm \: Work\; Done =60000 \: joule}$

[The unit'll be joule since N×M = J]

b) Potential energy stored in the water

$\rm \: P.E = m \cdot g \cdot h$

m = 100 kg
g = 10N/kg
h = 60

$\rm \: P.E =100 \:kg \: \times \cfrac{10 \: N}{kg} \times 60$

$\boxed{\rm \: P.E =60000 \: joule}$

same condition here as well, N×M = J
c) Power of the Pump

$\rm \: P = W/T$

where P = Power; W = Work done & T = Time taken
As we got the value of work done on question (a),& ATQ time taken is 20 S.

$\rm \: P = \cfrac{60000 \: joule}{20 \: seconds} =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}$

d) Power rating of the pump = 3 kW

Assumption: The pump is 100% efficient & works well.

6 0

2 years ago

5. A projectile is fired in Earth's gravitational field with a horizontal velocity of y = 9.00 m/s.

aniked [119]

Answer:

<h3>4.13m</h3>

Explanation:

Given

Horizontal velocity = 9.00m/s

time taken = 0.550 s

Required

How far does the projectile fall in the vertical direction

Using the formula for finding the maximum height of the projectile

H = U²sin²θ/2g where;

U = 9.00m/s

θ = 90° (object launched in the vertical direction)

g = 9.81m/s²

Substituting the given parameters into the formula;

H = 9²sin²90/2(9.81)

H = 81(1)/19.62

H = 81/19.62

H = 4.128 m

H ≈ 4.13m

Hence the distance that the projectile fall in the vertical direction is 4.13m

3 0

4 years ago

(a) Calculate the electric field strength near a 10.0 cm diameter conducting sphere that has 1.00 C of excess charge on it. (b)

Romashka-Z-Leto [24]

Electric field strength near the surface of sphere is given by

$E = \frac{kQ}{r^2}$

Now we have

Q = 1 C

R = 5 cm

now electric field is given by

$E = \frac{9*10^9 * 1}{0.05^2}$

$E = 3.6 * 10^{12} N/C$

Part B)

Since charge on the sphere is very large to electric field near its surface is too high. So this is reasonable result near it.

Part c)

As per our assumption the charge is uniformly distributed on the surface and there is no effect of surrounding charge or surrounding electric field.

4 0

3 years ago