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Rzqust [24]
4 years ago
6

What is the answer to this question?

Physics
1 answer:
dem82 [27]4 years ago
3 0
Acceleration is a change in *speed* over time. In this case, the speed of the car increased by 90 km/hr in 6 s, giving it a rate of 90 km/hr/6s, or 15 km/hr/s. We’re asked for the acceleration in m/s^2, though, so we’ll need to do a few conversions to get our units straight.

There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:

(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2

We can reduce this to:

(15 x 10)m/36 s^2 = 150 m/36 s^2

Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
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I am an animal named after the animal that I eat, what am I?
Gnom [1K]

the answer is an anteater

7 0
4 years ago
Radioactive nuclei decay to become unstable. Is this true or false?
Damm [24]

The unstable nuclei undergo radioactive decay. The nucleus decay in form of emitting the radiations or changing into the different chemical element.

Thus, the nucleus decay takes place till the nuclei become stable.

Hence, given statement is false.

5 0
1 year ago
A 6 ft tall person walks away from a 10 ft lamppost at a constant rate of 5 ft/s. What is the rate (in ft/s) that the tip of the
nalin [4]

Answer:

12.5 ft/s

Explanation:

Height of person = 6 ft

height of lamp post = 10 ft

According to the question,

dx / dt = 5 ft/s

Let the rate of tip of the shadow moves away is dy/dt.

According to the diagram

10 / y = 6 / (y - x)

10 y - 10 x = 6 y

y = 2.5 x

Differentiate both sides with respect to t.

dy / dt = 2.5 dx / dt

dy / dt = 2.5 (5) = 12.5 ft /s

8 0
4 years ago
Time-outs are ineffective because
maria [59]
I think the correct answer would be the child quietly sits until the timer goes off. It is ineffective because the child would be used to this process and would come to a time where he will not get something from it.Also, it would make a child feel isolated and alone which would affect hi self-esteem in a long run.
8 0
3 years ago
Read 2 more answers
A child of mass M is swinging on a swing set. The ropes attaching the swing to the top bar have length L. Find the gravitational
myrzilka [38]

Answer:

(a) 0

(b) 10ML

(c) 10ML(1 - cos(\theta))

(d) 10ML(1 + sin(\phi))

Explanation:

(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.

(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be

E_h = mgh = 10ML

where g = 10m/s2 is the gravitational acceleration.

(c) At angle \theta from the vertical. Vertically speaking, the child should be at a distance of Lcos(\theta) to the swinging point, and a vertical distance of L - Lcos(\theta) to the lowest position. His potential energy to this point would be:

E_{\theta} = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))

(d) at angle \phi from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of Lsin(\phi) from the swinging point and higher than it. Therefore his vertical distance to the lowest point is L + Lsin(\phi) = L(1 + sin(\phi))

His potential energy to his point would be:

E_{\phi} = mgh = 10ML(1 + sin(\phi))

5 0
4 years ago
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