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s2008m [1.1K]
3 years ago
15

Help me it's due tomorrow

Physics
1 answer:
Tresset [83]3 years ago
4 0
I think it's B. :) <3

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A horizontal disk with a radius of 23 m rotates about a vertical axis through its center. The disk starts from rest and has a co
lys-0071 [83]

Answer:

time is 0.42 sec

Explanation:

Given data

radius = 23 m

angular acceleration = 5.7 rad/s²

to find out

time

solution

we know that radius is constant so that

tangential acceleration At = angular acceleration × radius   ............. 1

tangential acceleration =  5.7 × 23 = 131.1 m/s²

and

radial acceleration Ar =  (angular velocity)² × radius    ........................2

we consider angular velocity = ω

this is acting toward center

so

compare 1 and 2

At = Ar

5.7 r =ω³ r

ω = √5.7 = 2.38746 rad/s

so

ω = 5.7 t

2.387 = 5.7 t

t =  2.387 / 5.7

t = 0.4187

time is 0.42 sec

8 0
3 years ago
A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.
Nataliya [291]

Answer:

Approximately 1.2 \times 10^{4}\; {\rm J} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: g = 9.81\; {\rm N \cdot kg^{-1}}.

The change in the gravitational potential energy ({\rm GPE}) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass m increased by \Delta h, the {\rm GPE} of that object would have increased by m\, g\, \Delta h.

In this question, the height of this object increased by \Delta h = 6.0\; {\rm m}. The mass of this object is m = 200\; {\rm kg}. Thus, the {\rm GPE} of this object would have increased by:

\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}.

(Note that 1\; {\rm N \cdot m} = 1\; {\rm J}.)

3 0
2 years ago
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of th
Anon25 [30]

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

8 0
3 years ago
A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov
Inessa [10]

Answer:

K = ½ kg m2 / s2

Explanation:

The kinetic energy rotation is given by the expression

    K = ½ I w²

Where I is the moment of inertia and w is the angular velocity.

We have a relationship between linear and angular velocity

    v = w r  

Where r  the distance from the point of rotation, in this case it is the point or where the ball is, so the linear velocity is the same as the ball.

      w = v / r

We replace

      K = ½ I (v / r)²

Let's calculate

      K = ½ kg m2 )(m/s )/ m) 2

      K = ½ kg m2 / s2

7 0
3 years ago
(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the ca
gulaghasi [49]

Answer:

(a) E_{ c} = 112.5 \mu J

(b) E'_{ c} = 18 \mu J

Solution:

According to the question:

Capacitance, C = 1.00\mu F = 1.00\times 10^{- 6} F

Voltage of the battery, V_{b} = 15.0 V

(a)The Energy stored in the Capacitor is given by:

E_{c} = \frac{1}{2}CV_{b}^{2}

E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}

E_{ c} = 112.5 \mu J

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

E'_{c} = \frac{1}{2}CV'_{b}^{2}

E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}

E'_{ c} = 18 \mu J

6 0
4 years ago
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