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3 years ago

10

A mechanic pushes perpendicularly on the end of a 0.15m wrench. If the

1 answer:

kumpel [21]3 years ago

4 0

Answer:

τ = 34.5 N.m

Explanation:

The force applied by the mechanic is 230 N.

Thus, F = 230 N

Now, this force is applied in a counterclockwise manner while pushing perpendicularly at the end of the 0.15m long wrench.

Thus, r = 0.15 m

Formula for the torque in this case is;

τ = Fr sin θ

θ = 90° since mechanic pushed perpendicularly.

Thus;

τ = 230 × 0.15 × sin 90

τ = 34.5 N.m

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Write the equation expressing the relationship "y varies directly as x." use k as the constant of proportionality.

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The equation expressing the statement "y varies directly as x" is $y=kx.$

Explanation

The statement that $y$ varies directly as $x$ is analogous to saying that the ratio of $y$ to $x$ is constant. In other words, when $x$ increases, $y$ likewise increases and that when $x$ decreases, $y$ decreases proportionally.

Mathematically, we express the relationship that the ration of $y$ is to $x$ is constant is expressed as; $\frac{y}{x} =k$ where $k$ is the constant of proportionality.

We can then solve the relationship for $y$ to determine the correct form of the relationship as shown below,

$\frac{y}{x} =k\\\rightarrow y=kx$

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3 years ago

A man's higher initial acceleration means that a man can outrun a horse over a very short race. A simple - but plausible - model

Lera25 [3.4K]

Answer:

t_man = 10.16 s, t_horse = 10.73 s, the winner is the man

Explanation:

To solve this problem we are going to find the time of each one separately.

Man we look for distance and time during acceleration

x₁ = v₀ t₁ + ½ a₁ t₁²

as it comes out of rest its initial velocity is zero

x₁ = ½ a₁ t₁²

x₁ = ½ 6.0 1.8²

x₁ = 9.72 m

at this point its speed is

v₁ = v₀ + a t

v₁ = 0 + 6 1.8

v₁ = 10.8 m / s

From here on it continues at constant speed, the distance that the man needs to travel from the point where the man leaves at 100m is

x₂ = 100 - x₁

x₂ = 100- 9.72

x₂ = 90.28 m

the time for this part is

v₁ = x₂ / t₂

t₂ = x₂ / v₂

t₂ = 90.28 / 10.8

t₂ = 8.36 s

the total time for the man is

t_man = t₁ + t₂

t _man = 1.8 + 8.36

t_man = 10.16 s

We repeat the calculation for the horse

distance traveled during the acceleration period

x₃ = v₀ t + ½ a₂ t₃²

as part of rest its initial velocity is zero

x₃ = ½ a₂ t₃²

x₃ = ½ 5.0 4.8²

x₃ = 57.6 m

the velocity at this point is

v₃ = v₀ + a₂ t₃

v₃ = 0 + 5 4.8

v₃ = 24 m / s

the rest of the route is at constant speed, the remaining distance

x₄ = 200 - x₃

x₄ = 200 - 57.6

x₄ = 142.4 m

the time to go through it is

t₄ = x₄ / v₃

t₄ = 142.4 / 24

t₄ = 5.93 s

the total time for the horse is

t_horse = t₃ + t₄

t_horse = 4.8 + 5.93

t_horse = 10.73 s

when we compare the times we see that the man arrives a little before the horse, the winner is the man

5 0

3 years ago

A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.

Arturiano [62]

Answer:

a) $\alpha=0.5117\ rad.s^{-2}$

b) $\theta=8.4\ rad$

Explanation:

Given:

initial rotational speed of phonograph, $\omega_i=0\ rad.s^{-1}$

final rotational speed of phonograph, $N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}$

time taken for the acceleration, $t=5.73\ s$

a)

Now angular acceleration:

$\alpha=\frac{\omega_f-\omega_i}{t}$

$\alpha=\frac{2.932-0}{5.73}$

$\alpha=0.5117\ rad.s^{-2}$

b)

Using eq. of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2$

$\theta=8.4\ rad$

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3 years ago