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3 years ago

14

A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer the answer to the neares

t tenth.

1 answer:

loris [4]3 years ago

6 0

Answer:

speed = 7.9 m/s

Explanation:

speed = total distance / time taken

speed = 300 / 38

speed = 7.89473684 m/s

to the nearest tenth

speed = 7.9 m/s

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

3 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

frozen [14]

Answer:

0.285

Explanation:

Given two forces of different magnitude, it is important to note that the product of normal force and coefficient of kinetic friction should be equal to the sum of these two forces at equilibrium. Therefore, this can be Mathematically expressed as:

$N/\mu_k=F_1+F_2\\\\$

where N is normal force, $\mu$ is coefficient of static friction, F is force and subscripts 1 and 2 represent larger and smaller magnitude forces respectively. Making $\mu$ the subject of the formula then

$\mu_k=\frac{F_1+F_2}{N}$

Since normal force N is also given by mg where m is mass of object and g is acceleration due to gravity then substituting N with mg we obtain that

$\mu_k=\frac{F_1+F_2}{mg}$ and substituting the figures given in the question, taking g as 9.81 we obtain that

$\mu_k=\frac { 430 N+380 N}{290\times 9.81}=0.285$

Hence,the coefficient of kinetic energy is 0.285 as calculated

4 0

4 years ago

Water having a density of 1000 kg/m^3 is flowing with a velocity of 3 m/s through a round pipe. There is a restriction within th

Fofino [41]

Answer:

12 m/s

Explanation:

Using the continuity equation, which is an extension of the conservation of mass law

ρ₁A₁v₁ = ρ₂A₂v₂

where 1 and 2 indicate the conditions at two different points of flow, in this case, point 1 is any normal position in the pip and point 2 is the conditions at the restriction.

ρ = density of the fluid flowing; note that the density of the fluid flowing (water) is constant all through the fluid's flow

A₁ = Cross sectional Area of the pipe at point 1 = (πD₁²/4)

A₂ = Cross sectional Area of the pipe at the restriction = (πD₂²/4)

v₁ = velocity of the fluid flowing at point 1 = 3 m/s

v₂ = velocity of the fluid flowing at The restriction = ?

ρ₁A₁v₁ = ρ₂A₂v₂

Becomes

A₁v₁ = A₂v₂ (since ρ₁ = ρ₂)

(πD₁²/4) × 3 = (πD₂²/4) × v₂

3D₁² = D₂² × v₂

But

D₂ = (D₁/2)

And D₂² = (D₁²/4)

3D₁² = D₂² × v₂

3D₁² = (D₁²/4) × v₂

(D₁²/4) × v₂ = 3D₁²

v₂ = 4×3 = 12 m/s

8 0

3 years ago

If one doubles the emfs in a circuit and doubles the resistances in the circuit at the same time, what happens to the currents t

bija089 [108]

Answer:

Same

Explanation:

Let R be the resistance of the cell and E be the emf.

Let i be the current in the circuit.

i = E / R ..... (1)

Now emf and resistance be doubled.

so, i' = 2E/2R = E/R

So, i' = i (From equation (1)

Current remains same.

3 0

3 years ago

Precautions taken when doing experiment on simple pendulum

olchik [2.2K]

Answer:

1.)The bob of pendulum should be displaced with a small angle.

2) The amplitude of the oscillation of a simple pendulum should be small.

3) Fans should be switched off to reduce the air resistance.

4)The simple pendulum should be oscillate in a vertical plane

Hope this helped you

7 0

3 years ago