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Eva8 [605]
4 years ago
12

A runner starts from rest and accelerates uniformly to a speed of 8.0 meters per second in 4.0 seconds. the magnitude of the acc

eleration of the runner is
Physics
1 answer:
Paha777 [63]4 years ago
7 0
The magnitude of the acceleration of the runner is given by:
a= \frac{v_f-v_i}{t}
where
v_f=8.0 m/s is the final speed of the runner
v_i=0 is the initial speed of the runner
t=4.0 s is the time taken

By substituting data into the equation, we find the magnitude of the acceleration:
a= \frac{8.0 m/s -0}{4.0 s}=2 m/s^2
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The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the d
tamaranim1 [39]

Answer:

26852726.19\ \text{N}

57.52^{\circ}

Explanation:

r = Radius of circle = 7 m

w = Width of dam = 60 m

h = Height of the dam will be half the radius = \dfrac{r}{2}

A = Area = rw

V = Volume = w\dfrac{\pi r^2}{4}

Horizontal force is given by

F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times  7\times 60\\\Rightarrow F_x=14420700\ \text{N}

Vertical force is given by

F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}

Resultant force is

F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}

The hydrostatic force on the dam is 26852726.19\ \text{N}.

The direction is given by

\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}

The line of action is 57.52^{\circ}.

7 0
3 years ago
PLEASE HELP!!!!! which statements correctly conpare the masses of protons,neutrons,and electrons​
Anastasy [175]
Protons and neutrons have similar mass
Electrons are smaller then a proton or a neutron
8 0
3 years ago
Two pebbles, Pebble A and Pebble B, are thrown horizontally with the same force. Pebble A's mass is 3 times the
Hitman42 [59]

Answer:

Pebble A has 1/3 the acceleration as pebble B.

Explanation:

F = m×a

mass of a = 3 × mass of b (m_a = 3 × m_b)

Same starting force, F

m_a = mass of a

m_b = mass of b

a_a = acceleration of a

a_b = acceleration of b

F = m_a × a_a = m_b × a_b

3 × m_b × a_a = m_b × a_b

3 × a_a = a_b

OR

a_a = a_b / 3

5 0
3 years ago
5. A hollow cylinder of mass m, radius Rc, and moment of inertia I = mRc2 is pushed against a spring (with spring constant k) co
Makovka662 [10]

Explanation:

(a) Draw a free body diagram of the cylinder at the top of the loop.  At the minimum speed, the normal force is 0, so the only force is weight pulling down.

Sum of forces in the centripetal direction:

∑F = ma

mg = mv²/RL

v = √(g RL)

(b) Energy is conserved.

EE = KE + RE + PE

½ kd² = ½ mv² + ½ Iω² + mgh

kd² = mv² + Iω² + 2mgh

kd² = mv² + (m RC²) ω² + 2mg (2 RL)

kd² = mv² + m RC²ω² + 4mg RL

kd² = mv² + mv² + 4mg RL

kd² = 2mv² + 4mg RL

kd² = 2m (v² + 2g RL)

d² = 2m (v² + 2g RL) / k

d = √[2m (v² + 2g RL) / k]

8 0
3 years ago
When water in a lake freezes, the ice that forms floats on top of any water that is still liquid. Why does the ice float?
ddd [48]

Answer:

it floats because the lake is cold at that moment so when part of the lake freezes it still remains solid and floats because of the lake and the surrounding of the lake is still cold

5 0
4 years ago
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