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3 years ago

A disc is thrown through the air for 1.5 min with a power output of 12.5 W. How much work is done when throwing the disc?

Physics

1 answer:

klasskru [66]3 years ago

7 0

Answer:

work = 1125 [J]

Explanation:

To solve this problem we must remember the definition of power, which is defined as the relationship between work and time. The power can be calculated using the following equation:

Power = work/time

Power = 12.5 [w]

work = joules [J]

time = 1.5 [min] = 90 [s]

work = 12.5*90

work = 1125 [J]

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Jonah rode his skateboard 350 meters down a straight road at a constant velocity. He skated

Marysya12 [62]

Answer:

250 m/min down the road

Explanation:

Velocity is equivalent to speed but it considers the direction of the object. Velocity is also calculated by dividing the distance travelled by time. Therefore, $v=\frac {d}{t}$ where d and t are distance and time respectively. Given that d is given as 350 m and t is 1.4 s then by substitution $v=\frac {350}{1.4}=250 m/min$ and the direction is down the road.

Velocity is 250 m/min down the road

8 1

2 years ago

A single slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum, when monochromatic

Margarita [4]

Answer:

$\lambda = 424.4 nm$

Explanation:

As we know by the formula of diffraction

$a sin\theta = N\lambda$

so we have

a = slit size

$\theta$ = angular position of Nth minimum

so we will have

for first minimum of 630 nm light

$a sin40 = 1(630 \times 10^{-9})$

$a = 9.8 \times 10^{-7} m$

Now for another wavelength second minimum is at 60 degree angle

$a sin60 = 2 \lambda$

$(9.8 \times 10^{-7}) sin60 = 2 \lambda$

$\lambda = 424.4 nm$

3 0

3 years ago

1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.

dimulka [17.4K]

Answer:

The acceleration of the crate is $0.3356\,\frac{m}{s^2}$

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

$F=m\,a$

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

$F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}$

7 0

2 years ago

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

Troyanec [42]

Answer:

The value is $\epsilon = 3.84 *10^{-5} \ V$

Explanation:

From the question we are told that

The diameter of the ring is $d = 18 \ mm = 0.018 \ m$

The length of the solenoid is $l = 25 \ cm = 0.25 \ m$

The diameter of the solenoid is $D = 5.0 \ cm = 0.05 \ m$

The number of turns is N = 1500

The change in current in the solenoid is $\Delta I = 20 \ A$

The time taken is $\Delta t = 1 \ s$

Generally the radius of the ring is

$r = \frac{d}{2}$

=> $r = \frac{0.018 }{2}$

=> $r = 0.009 \ m$

Generally the area of the ring is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

$\frac{dB }{dt} = \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^{-7} \ N/A^2$

$\frac{dB }{dt} = 4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}$

=> $\frac{dB }{dt} = 0.150816\ T/s$

$\epsilon = 0.150816 * 2.545 *10^{-4}$

=> $\epsilon = 3.84 *10^{-5} \ V$

3 0

2 years ago

our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three

Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

The mass of the left rock is $m_s = 2.25 \ kg$

The mass of the rock on the right $m_p = 10.1 kg$

The distance from fulcrum to the center of the pile of rocks is $r_p = 0.360 \ m$

Generally the torque produced by the pile of rock is mathematically represented as

$\tau = m_p * g * r_p$

Substituting values

$\tau = 10.1 * 9.8 * 0.360$

$\tau = 35.63\ N \cdot m$

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

The torque due to the single rock is

$\tau = m_s * g * r_s$

At equilibrium the both torque are equal

$35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

$r_s = \frac{35.63 }{m_s * g}$

Substituting values

$r_s = \frac{35.63 }{2.25 * 9.8}$

$r_s =1.62 \ m$

6 0

3 years ago