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3 years ago

PLS ANSWER FAST WILL MARK BRAINLIEST

Physics

1 answer:

densk [106]3 years ago

7 0

In position A

The mirror has to be observed in the ray reflected from the mirror.

Since the the angle of reflection is equal to the angle of incidence, the reflected ray must be in equal path as the incident ray.

That phenomenon is only in diagram A

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How much does it cost to leave a porch light on all night?

worty [1.4K]

Cost = (0.001) x (the wattage of the light) x (the number of hours it's left on) x (the cost of each kilowatt-hour of electrical energy where you live).

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3 years ago

The center of a long frictionless rod is pivoted at the origin and the rod is forced to rotate at a constant angular velocity Q

Doss [256]

Answer:

The answers can be found attached.

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3 years ago

A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t

Readme [11.4K]

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)

5 0

3 years ago

Activity 1 more or less, pls answer this ASAP

AURORKA [14]

Answer:

ididate is a good one and

5 0

3 years ago

A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

$\Sigma \tau = 0$

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
$\tau = r \times F$

Doing the summation using their respective lever arms:

$0 = L Tsin\theta - dF_g$

$dF_g = LTsin\theta$

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

$tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o$

Now, let's solve for 'T'.

$T = \frac{dMg}{Lsin\theta}$

Plugging in the values:
$T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}$

3 0

2 years ago