All categories

2 years ago

Acceleration Practice

Physics

2 answers:

Trava [24]2 years ago

8 0

Answer:

0.85 m/s²

Explanation:

Acceleration is change in velocity over change in time. In mathematically, it can be expressed as:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{\vec v_2 - \vec v_1}{t_2-t_1}}$

Our final velocity is given to be 17 m/s in 20 seconds. Initial velocity is at starting point which is 0 m/s in 0 second. Therefore:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17-0}{20-0}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17}{20}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = 0.85 \ \, \sf{m/s^2}}$

Therefore, the acceleration of a horse from starting point to 17 m/s in 20 seconds is 0.85 m/s²

just olya [345]2 years ago

4 0

Answer:

0.85 m/s^2

Explanation:

Acceleration = (final velocity - starting velocity) / time

Final velocity = 17

Starting velocity = 0

Time = 20

1. Substitute:

Acceleration = (starting velocity - final velocity) / time --> Acceleration = ( 17 - 0) / 20

2. Solve:

Acceleration = ( 17 - 0) / 20 -->

Acceleration = 0.85

You might be interested in

I didnt want my question public i made a mistake i want it taken down

mrs_skeptik [129]

Then report it and it might be taken down

5 0

2 years ago

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

3 years ago

You are removing branches from your roof after a big storm. You throw a branch horizontally from your roof, which is a height 3.

mart [117]

Answer:

The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s

Explanation:

The given parameters of the motion of the branch are;

The height from which the branch is thrown = 3.00 m

The horizontal distance the branch lands from where it was thrown, x = 8.00 m

The direction in which the branch is thrown = Horizontally

Therefore, the initial vertical velocity of the branch, $u_y$ = 0 m/s

The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;

s = $u_y$ ·t + 1/2·g·t²

Where;

$u_y$ = 0 m/s

s = The initial height of the object = 3.00 m

g = The acceleration due to gravity = 9.8 m/s²

∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²

t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246

The horizontal distance covered before the branch touches the ground, x = 8.00 m

Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;

uₓ = x/t = 8.00 m/((√30)/7 s)

Using a graphing calculator, we get;

uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s

The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.

3 0

3 years ago

Action reaction forces are in which one of newtons laws

Klio2033 [76]

Answer:

the third law

Explanation:

could u vote me brainliest plz? thx :)

4 0

3 years ago

A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor

Lorico [155]

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

7 0

3 years ago