The gravity on earth is 6 times greater than the gravity on the moon. An object weighing 2 lbs. on the moon will weigh:

Write the first equation of motion. Under what condition(s) is this equation valid?

Zepler [3.9K]

Explanation:

The first equation of motion in kinematics is given by :

$v=u+at$ .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

3 0

3 years ago

(1) Develop an equation that relates the rms voltage of a sine wave to its peak-to-peak voltage. a. If a sine wave has a peak-to

Aleks04 [339]

Answer:

(A) Equation will be $v=v_msin\omega t=0.75sin(18840t)$

(B) RMS value of voltage will be 0.530 volt

Explanation:

We have given peak to peak voltage of ac wave = 1.5 volt

Peak to peak voltage of ac wave is equal to 2 times of peak voltage

So $2v_{peak}=1.5volt$

$v_{peak}=\frac{1.5}{2}=0.75volt$

Frequency of ac wave is given f = 3 kHz

So angular frequency $\omega =2\pi f$ = 2×3.14×3000 = 18840 rad/sec

So expression of equation will be $v=v_msin\omega t=0.75sin(18840t)$ ( As phase difference is 0 )

Now we have to find the rms value of voltage

So rms voltage will be equal to $v_{rms}=\frac{v_{peak}}{\sqrt{2}}=\frac{0.75}{1.414}=0.530volt$

4 0

3 years ago

One problem with wind energy as a major source of electricity is _____. (1 point)

Alborosie

the problems that harnessing wind for energy is the expense of large tracts of land in populated areas.

hope this helps you

8 0

4 years ago

a bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s² for 10s. Find the distance moved by it during this interval o

BaLLatris [955]

Answer:

90 m

Explanation:

Acceleration, $a=\frac {v-u}{t}$ where v and u are final and initial velocities respectively, t is the time taken

Substituting $1 m/s^{2}$ for a, 4 m/s for u and 10 s for t then

1*10=v-4

v=14 m/s

From kinematic equations

$v^{2}=u^{2}+2as$

Making s the subject then

$s=\frac {v^{2}-u^{2}}{2a}=\frac {14^{2}-4^{2}}{2\times 1}=90 m$

6 0

3 years ago

A disc on a frictionless axle, starting from rest (0 rpm) can spin up to a rotation rate of 3820 rpm in a period of 2 seconds. (

Eddi Din [679]

Answer:

1000 Nm

2000 Nm

1.00007 seconds

Explanation:

I = Moment of inertia = 5 kgm²

$\alpha$ = Angular acceleration

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

t = Time taken

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=5\times 200\\\Rightarrow \tau=1000\ Nm$

The torque of the disc would be 1000 Nm

If $\alpha=400\ rad/s^2$

$\tau=I\alpha\\\Rightarrow \tau=5\times 400\\\Rightarrow \tau=2000\ Nm$

The torque of the disc would be 2000 Nm

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{3820\times \frac{2\pi}{60}-0}{400}\\\Rightarrow t=1.00007\ s$

It would take 1.00007 seconds to reach 3820 rpm

6 0

4 years ago