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NikAS [45]
2 years ago
15

During a race, a sprinter accelerated 1.8 m/s 2 in 2.5 seconds.How many meters per second did the sprint increase with this amou

nt of acceleration
Physics
1 answer:
trasher [3.6K]2 years ago
3 0

During a race, a sprinter accelerated 1.8 m/s 2 in 2.5 seconds. The sprint increase with this amount of acceleration by 4.5 m/s.

<h3>What is acceleration?</h3>

Acceleration is the time rate of change of velocity.

Acceleration a = velocity v / time t

1.8 = v/2.5

v = 4.5 m/s

The sprint increase with this amount of acceleration by 4.5 m/s.

Learn more about acceleration.

brainly.com/question/12550364

#SPJ1

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Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
3 years ago
A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr
kolbaska11 [484]

Answer:

0.002 N/C

Explanation:

Parameters given:

Charge of object, q = 5 mC = 5 * 10^{-3} C

Acceleration of object, a = 0.005 m/s^2

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> E = \frac{-ma}{q}

Solving for E, we have:

E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C

The magnitude will be:

|E| = |-0.002| N/C = 0.002 N/C

The electric field has a magnitude of 0.002 N/C.

4 0
3 years ago
A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0
3 years ago
Anyone going to be my friend
Artemon [7]

Explanation:

I'd love to but we cant talk right now cause its 12:22 am here and I'm gonna sleep now lol.

but let's follow each other.

who knows we might be able to help each other.

whaddya say?

have a good day ♡

5 0
2 years ago
A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
3 years ago
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