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skad [1K]
3 years ago
6

Plz help!!!

Physics
1 answer:
shtirl [24]3 years ago
3 0

Answer:

I think the answer to your question is true

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Starting from rest, a dragster travels a straight 1/4 mi racetrack in 7.10 s with constant acceleration. What is its velocity wh
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268.6567 mph  is its velocity when it crosses the finish line

d=(v1+v2 /2) x t

.25=(0+v2 /2) x 6.7/3600 hours

900=v2/2 x 6.7

v2=268.6567 mph as the speed with which the dragster crosses the finish

<h3>When acceleration is not zero, can speed remain constant?</h3>

The answer is that an accelerated motion can have a constant speed. Consider a particle travelling uniformly around a circle; it experiences acceleration since the motion's direction is changing, but it maintains a constant speed along the tangential axis throughout the motion.

Acceleration is the frequency of a change in velocity. Acceleration is a vector with magnitude and direction, much as velocity. For instance, if a car is moving in a straight path and speeding up, it is said to have forward (positive) acceleration, and if it is slowing down, it is said to have backward (negative) acceleration.

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1 year ago
How Many Negative (-) Electrons are there
makvit [3.9K]

Answer:

All electrons are negative(-) charged

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3 years ago
I am thinking of an unreactive nonmetal and Aisha suggested I use bromine. Is this a good choice? Explain your reasoning. If not
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3 years ago
A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does
Iteru [2.4K]

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

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6 0
2 years ago
A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel
Vesna [10]

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

3 0
3 years ago
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