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alex41 [277]
2 years ago
12

A hungry lion is looking for food. It went 5 miles (mi) before it spotted a zebra to stalk. It took this lion 0.2 hours to get t

o the zebra. What is the lion’s speed?
Question 1 options:

1 mi


1 mi/h


25 mi


25 mi/h
Physics
1 answer:
Helen [10]2 years ago
5 0

To solve this exercise, the following speed equation must be applied:

 V = d / t (mi/hr)

 V: speed (mi/hr)

 d: distance (mi)

 t: time (s)

We know the distance (d= 5 miles) and the time (t= 0.2 hours), so we proceed to replace these values in the formula to calculate the lion’s speed:

 V = d / t

 V = 5 mi/0.2 hr

 V = 25 mi/hr

 The lion’s speed is 25 mi/hr, so the last option is the right answer

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Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.
galina1969 [7]

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}

Here, k is the Coulomb's constant whose value is

k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the magnitude of charge will be

\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.

6 0
1 year ago
A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?
Artist 52 [7]

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f =  μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

6 0
3 years ago
Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert
damaskus [11]

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

F = k \frac{q\cdot q}{r^2}

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

q+\frac{q}{2}=\frac{3}{2}q

while the other charge will be

q-\frac{q}{2}=\frac{q}{2}

So, the new force will be

F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F

So, the force will decrease to 3/4 of its original value.

6 0
3 years ago
A skateboarder drops in off the top of one side of the half pipe shown below. She does not push off and starts from rest. She st
solong [7]

Answer:

v

Explanation:

4 0
2 years ago
Two identical point charges are 3.00 cm apart. find the charge on each of them if the force or repulsion is 4.00 x 10^-7. (Use C
DanielleElmas [232]

Answer:

Charge on each is 2 x 10⁻¹⁰.

Explanation:

We know that Force between two point charges is given b the Coulomb's law as:

F = kq₁q₂/r^2

k = 9 x 10^9

r = 3.00 cm

= 0.03 m

q₁ = q₂

F = 4.00 x 10^-7

Rearranging the formula, we get:

F = k q²/r²

q² = Fr²/k

q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)

q² = 4 x 10⁻²⁰

q = 2 x 10⁻¹⁰

As there is force of repulsion between the charges, the charges must be both positive or both negative.

3 0
3 years ago
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