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castortr0y [4]
3 years ago
5

Send help my teacher is playing call me maybe

Physics
1 answer:
guajiro [1.7K]3 years ago
3 0

Answer:

lol

Explanation:

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A dock worker applies a constant horizontal force of 81.0 N to a block of ice on a smooth horizontal floor. The frictional force
AnnZ [28]

Answer:

The correct answer is:

(a) 84.240 kg

(b) 24.038 m

Explanation:

The given values are:

Force,

F = 81.0 N

Distance,

S = 13.0 m

Time,

t = 5.20 s

As we know,

The acceleration of mass will be:

⇒  a=\frac{2S}{t^2}

On substituting the given values, we get

⇒     =\frac{2\times 13.0}{(5.20)^2}

⇒     =\frac{26}{27.04}

⇒     =0.961538 \ m/s^2

(a)

The mass of the block will be:

⇒  m=\frac{F}{a}

On substituting the given values, we get

⇒       =\frac{81.0}{0.961538}

⇒       =84.240 \ kg

(b)

The final velocity after a given time i.e.,

t = 5.00 s

⇒  v=at

On substituting the values, we get

⇒     =0.961538\times 5.00

⇒     =4.8076 \ m/s

In time, t = 5.00 s

The distance moved by the block will be:

⇒  d=vt

On putting the values, we get

⇒     =4.8076\times 5.00

⇒     =24.038 \ m

5 0
3 years ago
How<br> are<br> earthquakes measured quantitatively
AlexFokin [52]

Answer:Intensity: The severity of earthquake shaking is assessed using a descriptive scale – the Modified Mercalli Intensity Scale. Magnitude: Earthquake size is a quantitative measure of the size of the earthquake at its source. The Richter Magnitude Scale measures the amount of seismic energy released by an earthquake.

Explanation:

4 0
3 years ago
The alignment of atoms where the north poles face one way and the south poles face the opposite way is also known as which of th
Pepsi [2]
The answer is B. Domain
When the atoms point to the same direction and aligned with one another, It will form a magnetic material in which  magnetization is in a uniform direction
This alignment usually happens stimunaleously  due to exchange interaction that caused by a response to an external field.
3 0
3 years ago
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
tekilochka [14]

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

m = 5925.4/3 kg

m = 1975.13 kg

So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

Learn more about balloons here:

brainly.com/question/21890581

8 0
3 years ago
A 23 g bullet is accelerated in a rifle barrel 62 cm long to a speed of 593 m/s. Use the work-energy theorem to find the average
Anuta_ua [19.1K]

Answer:

The average force exerted on the bullet while it is being accelerated is 6,522.52 N.

Explanation:

Given;

mass of the bullet, m = 23 g = 0.023 kg

length of the barrel, L = 62 cm = 0.62 m

speed of the bullet, v = 593 m/s

Applying work-energy theorem;

the work done in accelerating the bullet in the riffle = kinetic energy acquired by the bullet.

W = K.E

F x d = ¹/₂mv²

where;

d is the is the distance traveled by the bullet in the riffle = L

F(0.62) = ¹/₂ x 0.023 x (593²)

F(0.62) = 4043.964

F = (4043.964) / (0.62)

F = 6,522.52 N

Therefore, the average force exerted on the bullet while it is being accelerated is 6,522.52 N.

6 0
3 years ago
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