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Svet_ta [14]
3 years ago
10

What is the maximum speed with which a 1200-kg car can round a turn of radius 90.0 m on a flat road if the coefficient of fricti

on between tires and road is 0.80?
Physics
1 answer:
RideAnS [48]3 years ago
5 0

Answer:

Maximum speed of the car is 26.56 m/s.

Explanation:

Given that,

Mass of the car, m = 1200 kg

Radius of the curve, r = 90 m

The coefficient of friction between tires and the road is 0.8.

We need to find the maximum speed of the car. On the circular curve, the centripetal force is balanced by the force of friction. So,

\dfrac{v^2}{r}=\mu g\\\\v=\sqrt{\mu gr} \\\\v=\sqrt{0.8\times 9.8\times 90} \\\\v=26.56\ m/s

So, the maximum speed of the car is 26.56 m/s.

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A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,
Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

v_0 = 660 m/s

M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

v = 7934.2 m/s

5 0
3 years ago
Explain whether a tennis ball dropped from a high distance experiences an elastic collision or inelastic collision
hodyreva [135]
I need help on that too poop head
4 0
2 years ago
You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of
solong [7]

Answer:

The velocity of the man from the frame of  reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

                                V₂₁ = V₂ -V₁

∴                               V₂ =  V₂₁ + V₁

On substitution

                                 V₂ = 3 + 2

                                      = 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0
3 years ago
A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci
Simora [160]

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

4 0
2 years ago
Name one benefit and one limitation of Comparative investigations
GalinKa [24]
Comparative investigation is a research method that makes comparison across different countries of the world. Limitation : data set in different countries may be defined differently. Benefit : the method is flexible.
8 0
3 years ago
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