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mash [69]
2 years ago
13

Please help me! I don't understand how to do this!

Physics
1 answer:
dezoksy [38]2 years ago
3 0

From the information in the question, the acceleration of the cart is 2.55 ms-2.

We can find the acceleration using the following information from the question;

u = 0.9 m/s

v = 6 m/s

a =  2.0 seconds

But;

a = v - u/t

v = final velocity

u = initial velocity

t = time

a = 6 - 0.9/2 = 2.55 ms-2

Force= ma

m = mass

a = acceleration

F = 6 kg × 2.55 ms-2

F = 15.3 N

Learn more about acceleration: brainly.com/question/12134554

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3 years ago
What percent of air by volume is nitrogen​
nlexa [21]

Answer:

78 percent

Explanation:

I guess that's the right answer

5 0
3 years ago
A wire carrying a current of 26.9 A is bent into a circular arc with a radius of 0.6 cm that sweeps out 0.900 radians. What is t
melomori [17]

The magnetic field at the center of the arc is 4 × 10^(-4) T.

To find the answer, we need to know about the magnetic field due to a circular arc.

<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
  • According to Biot savert's law, magnetic field at the center of a circular arc is
  • B=(μ₀ I/4π)× (arc/radius²)
  • As arc is given as angle × radius, so

        B=( μ₀I/4π)×(angle/radius)

<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>

B=(μ₀ I/4π)× (0.9/0.006)

  = (10^(-7)× 26.9)× (0.9/0.006)

  = 4 × 10^(-4) T

Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.

Learn more about the magnetic field of a circular arc here:

brainly.com/question/15259752

#SPJ4

5 0
2 years ago
In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr
spin [16.1K]

Answer:

a) v_p=9.35m/s

Explanation:

From the question we are told that:

Open cart of mass   M_o=50.0 kg

Speed of cart   V=5.00m/s

Mass of package   M_p=15.0kg

Speed of package at end of chute V_c=3.00m/s

Angle of inclination   \angle =37

Distance of chute from bottom of cart   d_x=4.00m

a)

Generally the equation for work energy theory is mathematically given by

  \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

Therefore

  \frac{1}{2}u^2+gh=\frac{1}{2}v_p^2

  v_p=\sqrt{2(\frac{1}{2}u^2+gh)}

  v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}

  v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}

  v_p=9.35m/s

4 0
2 years ago
A car weighs 1323 N. what is its mass?
Shalnov [3]
If the car is on the moon, its mass is about 817 kg.

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6 0
2 years ago
Read 2 more answers
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