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3 years ago

Can someone answer these physics questions

Physics

1 answer:

USPshnik [31]3 years ago

6 0

1. The object moves -25 meters in a period of 3 seconds, so its average velocity is (-25 m)/(3 s) = -25/3 m/s ≈ -8.3 m/s

2. Ms. Patil's average velocity is the displacement over the period divided by the period length: (8 mi)/(0.25 h) = 32 mi/h

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How much power is generated by doing 500 j of work in 30 seconds

navik [9.2K]

Power is energy/time where energy is in joules and time is in seconds. The units for power is watts, W.

P = 500J/30s = 16.67 W

Hope this helps!

3 0

3 years ago

The box now rests on a frictionless ramp angled at 15◦ . The mover pulls up on a rope attached to the box to move it up the incl

Nataly [62]

Answer:

F = 84.61 N

Explanation:

As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.

component of Force along the incline = F cos(23° - 15°) = F cos(8°)

component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)

Equating the above two components of forces will give the minimum Force required.

F cos(8°) = 33*9.81*sin(15°)

F = 33*9.81*sin(15°) / cos(8°) (calculate the value using a scientific calculator)

3 0

3 years ago

Amy walks 3 m from her desk to the teacher's desk. From the teacher's desk, she then walks 4 m in the opposite direction to the

Afina-wow [57]

Do you have pictures of the tables?

4 0

3 years ago

A curved road has a radius of 120 m and a cant angle of 48 degrees. What is the maximum speed to stay on the curve in the absenc

zalisa [80]

Answer:

Maximum speed, v = 36 m/s

Explanation:

Given that,

The radius of the curved road, r = 120 m

Road is at an angle of 48 degrees. We need to find the maximum speed of stay on the curve in the absence of friction. On a banked curve, the angle at which it is cant is given by :

$tan\theta=\dfrac{v^2}{rg}$

g is the acceleration due to gravity

$v=\sqrt{rg\ tan\theta}$

$v=\sqrt{120\times 9.8\times \ tan(48)}$

v = 36.13 m/s

v = 36 m/s

So, the maximum speed to stay on the curve in the absence of friction is 36 m/s. Hence, this is the required solution.

7 0

3 years ago

An elastic circular bar is fixed at one end and attached to a rubber grommet at the other end. The grommet functions as a torsio

Artist 52 [7]

Answer:

2.1 rad(anticlockwise).

Explanation:

So, we are given the following data or parameters or information in the question above:

=> "The torsional stiffness of the spring support is k = 50 N m/rad. "

=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"

=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"

Hence;

G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.

Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).

==> 0.15/20 (V - w) + θ = 0.

==> 0.15/20 (V - w ) = -θ.

Where V = k = 50 N m/rad

w = 183.3 θ.

Therefore, w + Vθ = 500 Nm.

==> 183.3 + 50 θ = 500 Nm.

= 6.3

Anticlockwise,

θ = 2.1 rad.

4 0

3 years ago