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4 years ago

A car is moving with a constant speed of 20. meters

1 answer:

8 0

So the question is what is the total distance travelled by a car that is moving with constant speed of v=20m/s in t=2 mins. First, lets convert t=2 mins to t= 120s. Since v=s/t, to get the distance wee need to get s. To do that we need to multiply both sides of the equation with t so: s=v*t. Now we input our numbers and get s=(20m/s)*(120s)=2400m. So the correct answer is (4) 2400m.

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A positively charged insulating rod is brought close to an object that is suspended by a string. If the object is repelled away

Trava [24]

Answer

A) Positively charged two insulating rod are brought closed to an object they repel each other. It means the Object is positively charged. Because similar charge repel each other.

The correct answer is Option A.

B) we know force between to charges is calculated using Formula

$F = \dfrac{kQ_1Q_2}{r^2}$ .......(1)

form the given condition in the question

$F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r}{2})^2}$

$F' = \dfrac{k\dfrac{Q_1}{4}\dfrac{Q_2}{4}}{(\dfrac{r^2}{4})}$

$F' = \dfrac{4}{16}\dfrac{kQ_1Q_2}{r^2}$

from equation (1)

$F' = \dfrac{F}{4}$

hence, the correct answer is Option C.

3 0

3 years ago

3. Infiltration is the process by which rainwater becomes

amid [387]

Your answer is :

A.

Groundwater.

4 0

4 years ago

. (Serway 9th ed., 7-33) A 0.600-kg particle has a speed of 2.00 m/s at point A and a kinetic energy of 7.50 J at point B. What

k0ka [10]

Answer:

a) KA = 1.2 J

b) vB = 5.00 m/s

c) W = 6.30 J

Explanation:

m = 0.600 kg

vA = 2.00 m/s

KB = 7.50 J

a) KA = ?

b) vB = ?

c) W = ?

We can apply the folowing equations

K = 0.5*m*v²

and

W = ΔK = KB- KA

then

a) KA = 0.5*m*vA² = 0.5*(0.600 kg)*(2.00 m/s)²

⇒ KA = 1.2 J

b) KB = 0.5*m*vB² ⇒ vB = √(2*KB / m)

⇒ vB = √(2*7.50 J / 0.600 kg)

⇒ vB = 5.00 m/s

c) W = ΔK = KB- KA = (7.50 J) - (1.2 J)

⇒ W = 6.30 J

5 0

4 years ago

A garden hose with a small puncture is stretched horizontally along the ground. The hose is attached to an open water faucet at

tresset_1 [31]

Answer:

The pressure inside the hose 7000 Pa to the nearest 1000 Pa.

$h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}$

$V_{1} =V_{2}$

The pressure at the site of the puncture is $P₂ =7161.3 Pa$

Explanation:

According to Poiseuille's law, $P_{1} - P_{2} = \frac{128μLQ}{πD^{4} }$

Where $P_{1}$ is the pressure at a point $1$ before the leak, $P_{2}$ is the pressure at the point of the leak $2$ , μ = dynamic viscosity, L = the distance between points $1$ and $2$ , Q = flow rate, D = the diameter of the garden hose.