a car has a mass of 1.00 x 10^3 kilograms and it has an acceleration of 4.5 meters/second what is the net force on the car

The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2

dlinn [17]

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

$P=\dfrac{V^2}{R}$

Substitute all the values,

$P=\dfrac{3^2}{2}\\\\P=4.5\ W$

So, the power of this circuit is equal to 4.5 watts.

5 0

3 years ago

NEED ANSWERS ASAP !!!

morpeh [17]

Answer:

I think its B

Explanation:

because "This means that when you rubbed the plastic comb along your hair, your hair resisted the movement of the comb and slowed it down. The friction between two surfaces can cause electrons to be transferred from one surface to the other."

3 0

3 years ago

Read 2 more answers

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

7 0

2 years ago

A motorcycle has 100,000 J of kinetic energy and is traveling at 20 m/s. Find its mass.

tankabanditka [31]

Answer:

<h3>The answer is 500 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{2KE}{ {v}^{2} } \\$

v is the velocity

KE is the kinetic energy

From the question we have

$m = \frac{2 \times 100000}{ {20}^{2} } = \frac{200000 }{400} = \frac{2000}{4} \\$

We have the final answer as

Hope this helps you

7 0

3 years ago

A piece of plastic with a mass of 15 g is

satela [25.4K]

Answer:

The density of plastic is equal to 0.6 g/mL.

Explanation:

Given that,

The mass of piece of plastic, m = 15 g

It is placed in a graduated cylinder. The water level in the graduated cylinder rises from 30 mL to 55 mL when the plastic is added.

We need to find the density of plastic.

Rise in volume = 55 mL - 30 mL

= 25 mL

The density of an object is given by :

$d=\dfrac{m}{V}\\\\d=\dfrac{15\ g}{25\ mL}\\\\d=0.6\ g/mL$

So, the density of plastic is equal to 0.6 g/mL.

5 0

3 years ago