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Anton [14]
2 years ago
13

If a car travels 60km/hr, how long would it take the car to travel 300km? *

Physics
2 answers:
lina2011 [118]2 years ago
4 0

Answer:

5  hurs

Explanation:

Irina-Kira [14]2 years ago
3 0

Answer:

5 hours

Explanation:

A way you could solve this is:

300 / 60 = 5

since the time measure is Kilometers per hour the answer should be in hour form.

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How does space promote science education
Fittoniya [83]
<span>NASA and the Mad Science Group of Montreal, Canada, have teamed in an effort to spark the imagination of children, encouraging more youth to pursue careers in science, technology, engineering and math. The two organizations recently signed a Space Act Agreement, officially launching the development of the Academy of Future Space Explorers.</span>
3 0
2 years ago
A heavy meterstick has a mass of 1 kg. When the meterstick is thrown like a spear past you, you measure its momentum to be 2mv.
CaHeK987 [17]

Answer:

Its length is measured to be 0.5 m

Explanation:

From theory of relativity (mass variation), we know that:

m = mo/√(1-v²/c²)

Where, m = relative mass

and, mo = rest mass

The momentum of stick while moving, will be:

P = mv

but, it is given in the form of rest mass as:

P = 2(mo)v

thus, by comparison;

2(mo)v = mv

using value of m from theory of relativity;

2(mo)v = (mo)v/√(1-v²/c²)

√(1-v²/c²) = 1/2     ______ eqn(1)

Now, for relativistic length (L), we have the formula from same theory of relativity;

L = (Lo)√(1-v²/c²)

The rest length (Lo) of meter stick is 1 m, and the remaining term on right side √(1-v²/c²), known as Lorentz Factor, can be given by eqn (1), as equal to 1/2.

Thus,

L = (1 m)(1/2)

<u>L = 0.5 m</u>

4 0
3 years ago
Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha
Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N.  Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.  

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation  is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

a_B = a_A-0.5 \frac{m}{s^2}

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0
3 years ago
Read 2 more answers
Why would you have trouble breathing at high altitudes?
monitta

Answer:

A. It is colder at the top of a mountain

Explanation:

4 0
2 years ago
What are tiny sacs at the end of the bronchioles filled with air called?
WARRIOR [948]

Answer:

Alveoli

The bronchioles end in tiny air sacs called alveoli, where oxygen is transferred from the inhaled air to the blood.

Hope this helps :)

7 0
3 years ago
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