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2 years ago

Please help I’ll give brainliest

Physics

1 answer:

In-s [12.5K]2 years ago

4 0

Answer:

2nd option

Explanation:

Static electricity the charges are at rest and they are accumulating on the surface of the insulator.

Current electricity the electrons are moving inside the conductor.

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Why would genetic material need to be able to reproduce

Lynna [10]

So that the next generation could inherit the previous adaptations and instinct and be similar to the parents. if that didn't happen, a cat might give birth to a giraffe

7 0

3 years ago

A car travels 40 miles in 30 minutes.

lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 $m/s^2$

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles = 40 x 1.60934

so 40 miles = 64.3738 Km

similarly converting 30 minutes into hours

1 minute = $\frac{1}{60}hours$

30 minute = $\frac{30}{60}hours$

30 minute = $\frac{1}{2}hours$

Now

Average velocity = $\frac{Speed}{time}$

Substituting Values,

Average velocity = $\frac{64.3738}{\frac[1}{2}}$

Average velocity = $64.3738 \times 2$

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting velocity to m/s

$128.74 \times \frac{5}{18}$

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy K= $\frac{1}{2}mv^2$

Substituting the values,

K= $\frac{1}{2}1500(35.75)^2$

K= $\frac{1}{2}1500(1278.06)$

K= $\frac{1500 \times (1278.06)}{2}$

K= $\frac{1917093.75}{2}$

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

$S= ut+\frac{1}{2}at^2$

Substituting the known values,

$s= ut+\frac{1}{2}at^2$

$s= (37.75)(15)+\frac{1}{2}a(15)^2$

$s=566.25+\frac{1}{2}a(225)$

$s=566.25+\frac{(225a)}{2}$ -------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

$a = \frac{v - u}{t}$

Substituting the values

$a = \frac{0 - 35.75}{15}$

$a = \frac{- 35.75}{15}$

a= - 2.38 $m/s^2$ ----------------------------------------(4)

Now substituting 4 in 3 we get

$s=566.25+\frac{(225( - 2.38)}{2}$

$s=566.25+\frac{-535.5}{2}$

$s=536.25-267.75$

s=268.8m--------------------------------------------------------------(5)

4 0

2 years ago

6. A light ray strikes a reflective plane surface at an angle of 560 with the surface.

Zolol [24]

Answer:

deez nouts

Explanation:

5 0

2 years ago

You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient

iren [92.7K]

Answer:

angle minimum θ = 41.3º

Explanation:

For this exercise let's use Newton's second law in the condition of static equilibrium

N - W = 0

N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0

the friction force formula is

fr = μ N

fr = μ W

we substitute

μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0

μ sin θ - cos θ + ½ cos θ= 0

μ sin θ - ½ cos θ = 0

sin θ / cos θ = 1/2 μ

tan θ = 1/2 μ

θ = tan⁻¹ (1 / 2μ)

θ = tan⁻¹ (1 (2 0.57))

θ = 41.3º

7 0

2 years ago

A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini

Volgvan

Answer:

The horizontal range will be $2.55\times 10^5m$

Explanation:

We have given initial speed of the shell u = $1.6\times 10^3m/sec$

Angle of projection = 51°

Acceleration due to gravity $g=9.8m/sec^2$

We have to find maximum range

Horizontal range in projectile motion is given by

$R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m$

So the horizontal range will be $2.55\times 10^5m$

6 0

2 years ago