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Cloud [144]
4 years ago
10

A 200-kg object has a 50-Newton rightward net force being applied to it. What is the magnitude of the rightward acceleration on

the object m/s^2?
Physics
1 answer:
katovenus [111]4 years ago
8 0
The formula for acceleration is force divided by mass.

Force = 50 N

Mass = 200 kg

Acceleration = ???

\frac{50}{200} = \frac{1}{4}

\frac{1}{4} = 0.25

~~

Acceleration = 0.25 m/s²

~~

Next time use, f = ma in a triangle,

F

M A

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
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Jason drives uphill from home to school, averaging 20 miles per hour. On the way back, he drives back downhill from school to ho
r-ruslan [8.4K]

Answer:

Average velocity = 30 miles per hour

Explanation:

Given that

Velocity for uphill = 20 miles per hour

Velocity for downhill = 60 miles per hour

We know that

Distance = Velocity x time

Total distance = Average velocity x total time

Lets take distance from home to school is s

Time taken from home to school= s/20 hr

Time taken from school to home= s/60 hr

Total time =s/20+s/60

Total distance = Average velocity x total time

s + s =  Average velocity x (s/20+s/60)

So average velocity = 30 miles per hour

6 0
3 years ago
A block of weight 45.7 N is hanging from a rope. The tension from the rope is pulling upward on the block. The block is accelera
viktelen [127]
<h2>Answer:</h2><h2></h2>

52.555 N

<h2>Explanation:</h2>

Let's use Newton's second law of motion here which states that the resultant force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) due to this force. i.e

∑F = m x a            ---------------------(i)

<em>Now, let's get the resultant force;</em>

Two main forces are acting on the rope;

i. the weight (W) of the block acting downwards.

Where;

W = mass of block(m) x gravity(g) = m x g

ii. the tension (T) in the rope acting upwards.

Therefore, the resultant force is the vector sum of these two forces as follows;

∑F = - W + T            [upward motion is taken as positive. hence -W and +T]

<em>Substitute ∑F = - W + T into equation (i) as follows;</em>

- W + T = m x a      ---------------------(ii)

<em>From the question;</em>

* Weight (W) of the block = 45.7N

=> mass (m) of the block = W / g = 45.7 / 10               [Taking g = 10m/s²]

=> m = 4.57 kg

* acceleration (a) = 1.50m/s²

<em>Substitute these values into equation (ii) as follows;</em>

- 45.7 + T = 4.57 x 1.50

- 45.7 + T = 6.855

<em>Solve for T;</em>

T = 6.855 + 45.7

T = 52.555 N

Therefore, the tension in the rope is 52.555 N

   

6 0
4 years ago
A 70 kg football player running at 3m/s north tackles an 80kg player running at 1.5 m/s south. what is the magnitude and directi
dimulka [17.4K]
It could never actually happen like this, but the question is
looking for you to 'conserve' the momentum.

Momentum of a moving object is (mass) x (velocity).
Like velocity, momentum has a direction.
Momentum is one of those things that's 'conserved'. 
That means that momentum can't appear out of nowhere, and
it doesn't disappear.  The total after the collision is the same as
the total was before the collision.  

Momentum of the skinny player:

                     (70 kg) x (3 m/s north) = 210 kg-m/s north.

Momentum of the heavy player:

                     (80 kg) x (1.5 m/s south)  =  120 kg-m/s south .

Total momentum before the collision is

                     (210 kg-m/s north) + (120 kg-m/s south)

                 =        90 kg-m/s north  .

It has to be the same after the collision.

                       (mass) x (velocity)  =  90 kg-m/s north.

The  mass after the collision is 150 kg, because they get
tangled up and stuck together, and they move together.

                                (150 kg) x (velocity)  =  90 kg-m/s north .

Divide each side
by  150 kg :                   velocity  =  (90 kg-m/s north) / (150 kg)

                                                =  (90/150)  (kg-m/s / kg  north)

                                                 =        0.6 m/s  north  .
3 0
3 years ago
The weight of the mass added to the hanger is equal to the extra force on the gas, but what area should we use to calculate the
kari74 [83]

Answer: according to the Avagadro's law, volume is directly propotional to no of moles: VXn

according to the Charles law, volume is directly propotional to  temperatue: VXT

according to the Boyle's law, volume is inversely propotional to P: VX1/P

when we combine them we get:

VXnT1/P

V=knT/P

k= R(universal gas constant)

V=RnT/P

PV=nRT  

8 0
3 years ago
What happens at a convergent boundary?
Julli [10]
I am not 100% sure on this, but I would have to say A or D. I am sorry I could not give you a definitive answer but there you go!
5 0
4 years ago
Read 2 more answers
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