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Effectus [21]
3 years ago
5

PLEASE HELP DUE BEFORE 11:30 TODAY!!!!

Physics
1 answer:
user100 [1]3 years ago
4 0

Answer:

D is not the a vector quantities

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Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be
miv72 [106K]

consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

h = 44.1 m

4 0
3 years ago
An elastic conducting material is stretched into a circular loop of 9.65 cm radius. It is placed with its plane perpendicular to
Nadya [2.5K]

Answer:

The induced emf in the coil is 0.522 volts.                        

Explanation:

Given that,

Radius of the circular loop, r = 9.65 cm

It is placed with its plane perpendicular to a uniform 1.14 T magnetic field.

The radius of the loop starts to shrink at an instantaneous rate of 75.6 cm/s , \dfrac{dr}{dt}=-0.756\ m/s

Due to the shrinking of radius of the loop, an emf induced in it. It is given by :

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=B\dfrac{-d(\pi r^2)}{dt}\\\\\epsilon=2\pi rB\dfrac{dr}{dt}\\\\\epsilon=2\pi \times 9.65\times 10^{-2}\times 1.14\times 0.756\\\\\epsilon=0.522\ V

So, the induced emf in the coil is 0.522 volts.                                

8 0
3 years ago
Please help on this one?
qwelly [4]

Answer:

D. An image that is smaller than the object and is behind the mirror

5 0
4 years ago
Students are experimenting with circuits in their physics class and they build the two working circuits pictured below. The batt
bija089 [108]

The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs.  In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases.  In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.

<h3>What is power of the circuit?</h3>

The power of the bulb or any resistor is equal to the product of voltage and  current flowing through it.

P = VI

Circuit A has bulbs in series while the circuit B has bulbs in parallel.

When bulb 3 added to circuit A,  the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.

The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.

Thus, the last option is correct.

Learn more about power.

brainly.com/question/2933971

#SPJ1

4 0
2 years ago
What type of plate boundary is shown in this illustration?
Inga [223]
This is a transform Boundrary
8 0
3 years ago
Read 2 more answers
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