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irina [24]
3 years ago
8

2. The car slows down from 36 m/s to 15 m/s over a 3.0 second time interval. What is its average acceleration?

Physics
2 answers:
stira [4]3 years ago
7 0

Answer:

wendys

Explanation:nutz go in yo mouth

VMariaS [17]3 years ago
6 0

Answer:

9 meter per second square

Explanation:

acceleration equation is final minus initial velocity divided by time in seconds.

36-15/3=9 meter per second square.

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The magnitude of the gravitational field strength near Earth's surface is represented by
Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

M - Mass of the planet Earth, measured in kilograms.

m - Mass of the person, measured in kilograms.

r - Radius of the Earth, measured in meters.

G - Gravitational constant, measured in \frac{m^{3}}{kg\cdot s^{2}}.

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

F = m \cdot g

Where:

m - Mass of the person, measured in kilograms.

g - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

g = \frac{G\cdot M}{r^{2}}

Given that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972 \times 10^{24}\,kg and r = 6.371 \times 10^{6}\,m, the magnitude of the gravitational field near Earth's surface is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

g \approx 9.82\,\frac{m}{s^{2}}

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

4 0
3 years ago
C. is what percent of 125?​
solmaris [256]

Answer:

Step 1: We make the assumption that 125 is 100% since it is our output value.

Step 2: We next represent the value we seek with $x$.

Step 3: From step 1, it follows that $100\%=125$.

Step 4: In the same vein, $x\%=125$.

Step 5: This gives us a pair of simple equations:

$100\%=125(1)$.

Explanation:

4 0
3 years ago
when in object moves stop moving changes speed or changes direction how do scientist describe that condition?
Kruka [31]

im taking test rn nd this question was on there, im saying unbalanced force, if it not correct i will put right answer, but im pretty sure the answer is unblanced

6 0
3 years ago
Read 2 more answers
The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply
kykrilka [37]

Answer: Got It!

<em>Explanation:  </em>let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

0  0

4 0
3 years ago
a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b
SVEN [57.7K]

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

4 0
3 years ago
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