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zlopas [31]
3 years ago
12

This picture shows an example of what?

Physics
1 answer:
Wewaii [24]3 years ago
5 0
The picture shows an example of velocity
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Carlos gives a grocery cart a 60-N push. The cart has a mass of 40 kg. What is the cart's acceleration?
andre [41]
<h3>Answer:</h3>

1.5 m/s²

<h3>Explanation:</h3>

We are given;

Force as 60 N

Mass of the Cart as 40 kg

We are required to calculate the acceleration of the cart.

  • From the newton's second law of motion, the rate of change in momentum is directly proportional to the resultant force.
  • That is, F = ma , where m is the mass and a is the acceleration

Rearranging the formula we can calculate acceleration, a

a = F ÷ m

  = 60 N ÷ 40 kg

  = 1.5 m/s²

Therefore, the acceleration of the cart is 1.5 m/s²

3 0
3 years ago
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0
3 years ago
Read 2 more answers
Katelyn (55 kg) is practicing a drop jump in the biomechanics lab. She steps off a plyometrics box, lands on the force plate, an
suter [353]

Answer:

J = 357.5 kg*m/s

Explanation:

  • The impulse exerted on Katelyn when she was on the force plate, is equal to the change in her momentum, according to Newton's 2nd Law.
  • Assuming as the positive direction the upward direction (coincident with the positive y-axis) we can express the initial momentum as follows:

       p_{o} = m*v_{o} = 55 kg * (-3.0 m/s)  (1)

  • By the same token, the final momentum is as follows:

       p_{f} = m*v_{f} = 55 kg * (3.5 m/s)  (2)

  • As we have already said, the impulse J is just equal to the change in momentum, i.e., the difference between (2) and (1):

      J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)

5 0
3 years ago
What is the source of all magnetism?
ivanzaharov [21]
The answer is electric current
3 0
3 years ago
A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it
Orlov [11]

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

F= ma (\rightarrow Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

\rho = \frac{m}{V} \Rightarrow m = \rho V

Part A) The buoyant force acting on the balloon is given as

F_b = ma

As mass is equal to the density and Volume and acceleration equal to Gravity constant

F_b = \rho V g

F_b = 1.2*323*9.8

F_b = 3798.5

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

F_{net} = F_b -W

F_{net} = F_b -(mg+\rho_H Vg)

F_{net} = 3798.5-(9.8*225*9.8*0.179*323)

F_{net} = 1030N

PART C) The additional mass that can the balloon support in equilibrium is given as

F_{net} = m' g

m' =\frac{F_{net}}{g}

m' = \frac{1030}{9.8}

m' = 105Kg

4 0
4 years ago
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