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3 years ago

15

If the second ball has a mass of 2.4kg and a constant acceleration of a⃗ 2= 2.8m/s2 j^ , what must the mass of the first ball be

?

1 answer:

kompoz [17]3 years ago

8 0

Hope this helps you!

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4. The Earth is the only planet that contains storms. Tor F 5. Lightning is a form of energy made up of protons that move from c

Mumz [18]

Answer:

4. F

5. F

Explanation:

jupiter constantly has hurricanes and uranus has storms often too.

lightning can move many ways including from cloud to cloud and even from ground ton sky (seriously look it up it's cool)

6 0

3 years ago

Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal

elena-s [515]

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

W = F║d
where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

W = (87 N)(4.5 m)
W = 391.5 J

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

5 0

3 years ago

Read 2 more answers

What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s

Ilia_Sergeevich [38]

Speed of wave is given as

$v = 5 m/s$

Wavelength of the wave is given as

$\lambda = 20 m$

now from the formula of wave time period we can say

$speed = \frac{wavelength}{time period}$

$5 = \frac{20}{T}$

$T = \frac{20}{5}$

$T = 4 s$

so it will have time period of T = 4 s

7 0

3 years ago

I need the answer asap everyone have a good day bye

Slav-nsk [51]

Im pretty sure its A cuz is closer to the earth.

5 0

3 years ago

In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give

sveta [45]

Answer:

The current pass the $R_2$ is $I = 0.25 A$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The voltage is $V = 3V$

The first resistance is $R_1 = 7 \Omega$

The second resistance is $R_2 = 5 \Omega$

Since the resistors are connected in series their equivalent resistance is

$R_{eq} = R_1 +R_2$

Substituting values

$R_{eq} = 7 + 5$

$R_{eq} = 12 \Omega$

Since the resistance are connected in serie the current passing through the circuit is the same current passing through $R_2$ which is mathematically evaluated as

$I = \frac{V}{R_{eq}}$

Substituting values

$I = \frac{3}{12}$

$I = 0.25 A$

3 0

4 years ago