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2 years ago

6

A truck covers 40.0 m in 7.80 s while uniformly slowing down to a final velocity of 1.70 m/s. Please provide with explanation an

d correct answer.

1 answer:

Karolina [17]2 years ago

4 0

Answer:

a = -0.223 m/s

Explanation:

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Dmitry [639]

equal and opposite reaction.

5 0

3 years ago

One horsepower (hp) is the amount of power required to lift a 75-kg mass a vertical distance of 1 m in 1 s. What is 2 hp equival

Vladimir [108]

Answer:

1470 W

Explanation:

Power: This can be defined as the rate at which work is done or energy is used up. The S.I unit of power is Watt (W).

The expression for power is given as,

P = Energy/time

P = mgh/t ...................... Equation 1

Where P = power, m = mass, h = height, t = time, g = acceleration due to gravity.

Given: m = 75 kg, g =9.8 m/s², h = 1 m, t = 1 s.

Substitute into equation 1

P = (75×1×9.8)/1

P = 735 W.

From the above,

1 hp = 735 W

2 hp = (2×735) W

2 hp = 1470 W.

Hence 2 hp = 1470 W

8 0

3 years ago

A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th

Stolb23 [73]

Answer:

The current is $I_b = 400 \ A$

Explanation:

From the question we are told that

The length of the segment is $l = 2.50 \ m$

The current is $I_a = 1000 \ A$

The force felt is $F = 4.0 \ N$

The distance of the second wire is $d = 5.0 \ cm = 0.05 \ m$

Generally the current on the second wire is mathematically represented as

$I_b = \frac{2 \pi * r * F }{ l * \mu_o * I_a }$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4 \pi * 10^{-7} \ N/A^2$

=> $I_b = \frac{2 * 3.142 * 0.05 * 4 }{ 2.50 * 4\pi *10^{-7} * 1000 }$

=> $I_b = 400 \ A$

4 0

2 years ago

The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8

Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) $\tau = \dfrac{\Delta L}{\Delta t}$

$\tau = \dfrac{0.8-3}{1.5}$

τ = -1.467 N m

b) angle at which fly wheel will turn

$\theta= \omega t +\dfrac{1}{2}\alpha t^2$

$\theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2$

$\theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2$

θ = 20.35 rad

c) work done on the wheel

W = τ x θ

W = -1.467 x 20.35 rad

W = -29.86 J

d) average power of wheel

$P_{av} =-\dfrac{W}{t}$

$P_{av} =-\dfrac{(-29.86)}{1.5}$

$P_{av} =19.91\ W$

7 0

2 years ago

Read 2 more answers

write a one or two summary paragraph discussing this experiment and the results use tye following questions 1 according to your

Monica [59]

Answer:heed

Explanation:

heed knows how to divide.

8 0

2 years ago