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3 years ago

True or false ? Radiant energy is generated by a petunia on Tuesday.

Physics

2 answers:

Fudgin [204]3 years ago

7 0

It is true that radiant energy or is generated by a petunia

Leya [2.2K]3 years ago

4 0

The answer is true that Radiant energy is generated by a petunia on Tuesday

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A 26.4 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the ca

MA_775_DIABLO [31]

Answer:

The horizontal component of the force exerted by the hinge on the beam is 47.15 N.

Explanation:

Given data:

Weight of beam = 26.4 kg

Angle between the beam and the cable is 90°

Beam inclination with respect to horizontal with an angle, $\theta = 23.4\°$

<u>We need to find the horizontal component of the force exerted by the hinge on the beam.</u>

Solution:

Let 'L' be length of the beam, 'T' be tension in the cable , $F_{h}$ be horizontal component of force by the hinge, and $F_{v}$ be vertical component of force by the hinge.

Take counterclockwise torque as positive.

Let us find torques around the hinge.

Torque by tension is given as:

$\tau = T \times L$

Torque by the force of gravity is given as:

$\tau_g= m g \frac{L}{2}\times cos \theta$

Torques by $F_{h}$ and $F_{v}$ are 0 as they act on the hinge itself.

Now, for equilibrium, net torque about the hinge is 0. So,

$\tau-\tau_g=0$

$T L - m g \frac{L}{2}\times \cos(\theta) = 0$

Dividing both sides by 'L', we get:

$T - m \frac{g}{2}\times \cos \theta = 0$

$T=m \frac{g}{2} \times cos \theta$ --------------------(1)

As per question, the cable makes 90° with the horizontal.

So, the net horizontal force is also zero. Therefore,

$F_{h} -T cos(90- \theta) = 0$

$F_h - T sin(\theta) = 0$

$F_h = T sin(\theta)$ --------------------------(2)

Plug the value of 'T' from equation (1) into equation (2). This gives,

$F_{h} = m \frac{g}{2} \times cos \theta \times sin \theta$

$F_{h} = 26.4 \times \frac{9.8}{2} \times cos(23.4) \times sin(23.4)$

$F_{h} = 47.15\ N$

Therefore, the horizontal component of the force exerted by the hinge on the beam is 47.15 N.

3 0

3 years ago

Which tools would sara us to find an irregularly shaped objects mass and volume A. a meter stick , a ruler , and water

tester [92]

I think it's D. Usually, to find the volume of an irregularly shaped object, you put it in water with a labeled beaker to measure how much the water rises. The balance would be used to measure the mass in grams.

4 0

3 years ago

Select the correct answer?

yKpoI14uk [10]

Answer:

The answer is d i think so yea

Explanation:

hope u hae a really good day

8 0

3 years ago

Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

$T^2\alpha R^3\\T^2=kR^3.......................(1)$

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

$R_p=16R_e...............(5)$

Substituting equation (5) into (4), we obtain the following;

$T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}$

$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

$T_p=64*365.25\\T_p=23376days$

4 0

3 years ago

A car of m = 1200. kg collides with a tree while traveling 60.0 mph. The collision occurs over a time period of 0.0500 seconds.

MrRissso [65]

You may know linear momentum is given by P= mass.velocity. Initially car is moving with some velocity so you know initial momentum of the car. Finally it comes to rest i.e final momentum of the car is 0. According to Newton's second law : Force = change in momentum /time. Applying this you'll get answer as 642840N. Hope it helped you. Revert back to me if you have any questions. Please check out the calculation it might be wrong!

5 0

3 years ago

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