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lord [1]
3 years ago
7

Which property of an object is the result of gravity?

Physics
2 answers:
Dmitry [639]3 years ago
5 0

Answer:

A. Mass

Explanation:

kap26 [50]3 years ago
4 0

Answer:

Weight

Explanation:

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Two car collide in an intersection. The speed limit in that zone is 30 mph. The car (mass of 1250 kg) was going 17.4 m/s (38.9).
Musya8 [376]

Answer:

u₂ = 3.7 m/s

Explanation:

Here, we use the law of conservation of momentum, as follows:

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\

where,

m₁ = mass of the car = 1250 kg

m₂ = mass of the truck = 2020 kg

u₁ = initial speed of the car before collision = 17.4 m/s

u₂ = initial speed of the tuck before collision = ?

v₁ = final speed of the car after collision = 6.7 m/s

v₂ = final speed of the truck after collision = 10.3 m/s

Therefore,

(1250\ kg)(17.4\ m/s)+(2020\ kg)(u_2)=(1250\ kg)(6.7\ m/s)+(2020\ kg)(10.3\ m/s)\\\\(2020\ kg)(u_2) = 8375\ N.s + 20806\ N.s - 21750\ N.s\\\\u_2=\frac{7431\ N.s}{2020\ kg}

<u>u₂ = 3.7 m/s</u>

5 0
3 years ago
Nancy applies a force to a violin string by pulling her bow across it. The vibrations of the string produce sound waves, which c
ra1l [238]

Answer:

option a) sound is the right answer

6 0
3 years ago
Read 2 more answers
At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)
Mama L [17]

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, a=0.54\ m/s^2

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

t=\dfrac{v-u}{a}

t=\dfrac{28.88-25.27}{0.54}

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

5 0
3 years ago
Which of the following is responsible for the formation of sinkholes F. Eroding soil on Earths’s surface G. Ice that builds up o
monitta
Ans is <span>I. Flowing groundwater that dissolves rock , such as limestone</span>
3 0
3 years ago
Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea
Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Where,

L = Vibrating length string

T = Tension in the string

\mu = Linear mass density

At the same time we have the expression for the number of beats described as

n = |f_1-f_2|

Where

f_1 = First frequency

f_2 = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

f \propto \sqrt{T}

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

n = f_2-f_1

f_2 = n+f_1

Replacing 3/sfor n and 202Hz for f_1,

f_2 = 3/s + 202Hz

f_2 = 3/s(\frac{1Hz}{1/s})+202Hz

f_2 = 206Hz

The frequency of the tightened is 205Hz

7 0
3 years ago
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