A weightlifter lifts a 1,250 N barbell 2 m in 3 s. How much power was used to lift the barbell?

The planet Pluto has 1/500 the mass and 1/15 the radius of Earth. What is the value of g on the surface of Pluto?

ahrayia [7]

<u>c 2.4 m/s²</u> is the answer

5 0

2 years ago

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is $v = 3.79 *10^{5} \ m/s$

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 144 \ kV /m = 144*10^{3} \ V/m$

The magnetic field is $B = 0.38 \ T$

The force due to the electric field is mathematically represented as

$F_e = E * q$

and

The force due to the magnetic field is mathematically represented as

$F_b = q * v * B * sin(\theta )$

Now given that it is perpendicular , $\theta = 90$

=> $F_b = q * v * B * sin(90)$

=> $F_b = q * v * B$

Now given that it is not deflected it means that

$F_ e = F_b$

=> $q * E = q * v * B$

=> $v = \frac{E}{B }$

substituting values

$v = \frac{ 144 *10^{3}}{0.38 }$

$v = 3.79 *10^{5} \ m/s$

7 0

2 years ago

An object exhibits SHM with an angular frequency w = 4.0 s-1 and is released from its maximum displacement of A = 0.50 m at t =

vivado [14]

Explanation:

It is given that,

Angular frequency, $\omega=4\ s^{-1}$

Maximum displacement, A = 0.5 m at t = 0 s

We need to find the time at which it reaches its maximum speed. Firstly, we will find the maximum velocity of the object that is exhibiting SHM.

$v_{max}=A\times \omega$

$v_{max}=0.5\times 4$

$v_{max}=2\ m/s$ ............(1)

Acceleration of the object, $a=\omega^2A$

$a=4^2\times 0.5$

$a=8\ m/s^2$ ...............(2)

Using first equation of motion we can calculate the time taken to reach maximum speed.

$v=u+at$

$t=\dfrac{v-u}{a}$

$t=\dfrac{2-0}{8}$

t = 0.25 s

So, the object will take 0.25 seconds to reach its maximum speed. Hence, this is the required solution.

4 0

2 years ago

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

valina [46]

The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
$T=2 t = 2 \cdot 0.100 s = 0.200 s$
Which means that the frequency is
$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
$\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s$

In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
$v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s$

6 0

3 years ago

What will happen if we put a magnet in cold water

gayaneshka [121]

Answer:

cold water strengthens the magnetic attractive force.

Explanation:

The molecules within the magnet will move slower because they have less kinetic energy,so there is less vibration within the magnets molecules.

8 0

2 years ago