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nataly862011 [7]
2 years ago
11

If you ran 15 km/h for 20 min, how much distance would you cover?

Physics
1 answer:
nikitadnepr [17]2 years ago
4 0
Since we have 15 kilometers per hour, and we're looking for 20 minutes, let's set up proportions.
20/60 minutes = x/15
20/60 = 1/3, so let's leave that simplified.
1/3 = x/15
Look at the denominators, 3 to 15 is a factor of 5, so multiply the numerator by 5.
1 • 5 = 5, so you will cover 5 kilometers in 20 minutes.

I hope this helps!
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Explain how to find the acceleration of an object that has one-dimensional horizontal motion.
Nostrana [21]
Acceleration is how much the velocity changes within a period of time so,

Acceleration= is the change in velocity divided by change in time

your units will be m/s squared
7 0
2 years ago
If Mrs. Reichelt throws a chromebook, because it won't login correctly, with a force of 8N, and the chromebook accelerates at 5m
suter [353]

Answer:

1.6 kg

Step-by-step Solution:

Since Force = mass × acceleration we have:

F = 8N

a= 5 m/s^2

m = ?

By plugging the values above into F=ma we obtain:

F=ma\\\\8=m(5)\\\\\frac{8}{5}=\frac{m(5)}{5}\\\\m=\frac{8}{5}=1.6

Therefore, the Chromebook has a mass of 1.6 kilograms.



7 0
3 years ago
Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. What is the veloc
Nuetrik [128]

The velocity of B after elastic collision is 3.45m/s

This type of collision is an elastic collision and we can use a formula to solve this problem.

<h3>Elastic Collision</h3>

v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1 - m_2}{m_1 + m_2}u_2

The data given are;

  • m1 = 281kg
  • u1 = 2.82m/s
  • m2 = 209kg
  • u2 = -1.72m/s
  • v1 = ?

Let's substitute the values into the equation.

v_1 = \frac{2*281*2.82}{281+209} -\frac{281-209}{281+209}(-1.72)\\v_1 = 3.45m/s

From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.

Learn more about elastic collision here;

brainly.com/question/7694106

4 0
2 years ago
Four point charges, each of magnitude 2.38 µC, are placed at the corners of a square 75.2 cm on a side. If three of the charges
poizon [28]

Answer:

The Electric Force on Negative Charge is 2.968 N

Explanation:

charge on each corner, q = 2.38 micro coulomb

Side of square, a = 75.2 cm

Coulombic constant, K = 8.98755 x 10^9 Nm²/C²

sides of the square are A,B,C and D

and all sides of a square are equal so

AB = BC = CD = DA = 75.2 cm = 0.752 m

Diagonal, AC = BD = 1.414 x 0.752 = 1.06 m

Electric field at D due to charge at A

EA= Kq÷AB^2

= 8.98755×10^9 × 9.87×10^-6 ÷ 0.752^2

EA= 156863.82 N/C

Similarly Electric field at D due to charge C

EC=Kq÷CD^2

= 8.98755×10^9 ×9.87×10^-6 ÷ 0.752^2

EC= 156863.82 N/C

Electric field at D due to charge at BB

EB=Kq÷BD^2

EB=8.98755×10^9 × 9.87×10^-6 ÷ 1.06^2

EB=78949.01 N/C

Resolve the compoents

Ex = EA + EB cos 45

Ex = 156863.82 + 78949.01 x 0.707

Ex = 212689.2 N/C

Ey = EC + EB Sin 45

Ey = 156863.82 + 78949.01 x 0.707

Ey = 212689.2 N/C

The resultant electric field is

E = 1.414 x 212689.2 = 300787.95 N/C

the electric force on the negative charge is

F = q x E

F = 9.87 x 10^-6 x 300787.95

F = 2.968 N

7 0
1 year ago
Is O2 considered one atom or 2 atoms?
natulia [17]
O2 is considered 2 atoms because O2 is 2 oxygen atoms.

4 0
3 years ago
Read 2 more answers
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