All categories

3 years ago

An object is accelerating from rest at a rate of 3 m/s2.How long will it take the object to attain 18 m/s ? *

Physics

1 answer:

Temka [501]3 years ago

7 0

Initial velocity =0, a=3m/s2, final velocity =18m/s, 18=3t, t=6 sec

You might be interested in

Which forces are acting on the student and the skateboard in the instant in which they are pushing off the wall? (Select all tha

diamong [38]

E all of the answers above correlate to the student and his skateboard

7 0

2 years ago

Read 2 more answers

¿Cuál es la frecuencia de una ola con una velocidad de 14 m / s y una longitud de onda de 20 metros?

Margaret [11]

Responder:

<h2>0.7Hertz </h2>

Explicación:

Usando la fórmula para calcular la velocidad de onda que se expresa como se muestra.

Velocidad de una onda = frecuencia * longitud de onda

v = fλ

Dada la velocidad de onda = 14 m / sy longitud de onda = 20 metros

frecuencia f = v / λ

f = 14/20

f = 0.7Hertz

La frecuencia de la onda es de 0.7 Hertz.

7 0

3 years ago

Which tool would you use to measure how long it takes a toy car to go down a ramp?

Ratling [72]

The answer will be C, a stopwatch :)

4 0

3 years ago

Read 2 more answers

A scientist testing to see how a cat “always” lands on their feet drops a cat upside down out of a third story window at a heigh

makkiz [27]

<u>Answer: </u>

Cat has 2.02 seconds to right itself.

<u>Explanation: </u>

Initial height of cat from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of cat in vertical direction = 0 m/s, acceleration = acceleration due to gravity = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So, cat has 2.02 seconds to right itself.

7 0

3 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

4 0

3 years ago