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jekas [21]
3 years ago
8

I need help in this pls i really need a answer

Physics
1 answer:
IrinaK [193]3 years ago
3 0

Answer:

In terms of distance, average speed is 20 km/h

In terms of displacement, average speed is 0 km/h.

Explanation:

Total distance:

= (40.0 - 10.0) + (20.0 - 10.0) + (40.0 - 20.0) \\  = 30.0 + 10.0 + 20.0 \\  = 60.0 \: km

Total time is 3.0 hours

but:

average \: speed =  \frac{total \: distance}{total \: time}  \\

In terms of distance.

substitute:

average \: speed =  \frac{60.0}{3.0}  \\  \\  = 20 \:  {kmh}^{ - 1}

displacement = ( - 30.0) + 10 .0+ 20.0 \\  = 0

In terms of displacement:

speed =  \frac{0}{3}  \\  \\  = 0 \:  {kmh}^{ - 1}

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A 1,300 kg wrecking ball hits the building at 1.07 m/s2.
dezoksy [38]
So we know F=ma
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2 years ago
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A ball is dropped from a height of 20m. If its velocity increases uniformly at the rate 10m/s2 with what velocity and after what
belka [17]

Answer: 2 seconds

Explanation:

6 0
3 years ago
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An electric oven has a resistance of 201 ohms and a voltage of 220 V. How much current does it draw?
asambeis [7]

1.1 A. An electric oven with a resistance of 201Ω and a voltage of 220V drwa a current of 1.1 A.

The easiest way to solve this problem is using the Ohm's Law I = V/R.

An electric oven has R = 201Ω, and a drop of voltage V = 220v, solve using I = V/R:

I = 220V / 201Ω

I = 1.09 A ≅ 1.1 A

5 0
3 years ago
by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th
xenn [34]

Answer:

<em>The force is now 9 times the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}

\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}

\displaystyle F'=9k\frac{q_1q_2}{d^2}

\displaystyle F'=9F

The force is now 9 times the original force

8 0
2 years ago
A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on
Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

F_f = ma (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg

And we can fidn the acceleration by using the formula:

a=\frac{v-u}{t}

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0
3 years ago
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