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3 years ago

In an electric motor, electrical energy is converted into

Physics

1 answer:

N76 [4]3 years ago

5 0

Answer:

idrk im jus here for the points $\lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n$ ∈ω≠≠

Explanation:

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A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra

seraphim [82]

1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

The motion of the bullet is a projectile motion, which consists of two separate motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

$d=\frac{u^2 sin 2\theta}{g}$

where

u is the initial speed of the projectile

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

$\theta=60^{\circ}$ (angle)

Solving the equation, we find the horizontal range:

$d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m$

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where

$v_y$ is the vertical velocity of the bullet after having covered a vertical displacement of $s$

$u_y$ is the initial vertical velocity

$a =-g=$ is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

$u_y = u sin\theta$

Also, the maximum height s is reached when the vertical velocity becomes zero,

$v_y =0$

Substituting into the equation and re-arranging for s, we find the maximum height:

$s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

3 years ago

Why did Ptolemy and Galileo had different views about the solar system?

iVinArrow [24]

Galileo only saw the system through a scope

5 0

3 years ago

Read 2 more answers

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F

Marrrta [24]

Answer:

Part a)

$x = 0.4 m$

Part b)

$v_i = 11.7 m/s$

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2$

$x = 0.4 m$

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

$W_f = -7( \pi R)$

$W_f = -7(\pi \times 1.00)$

$W_f = -22 J$

work done against gravity is given as

$W_g = - mg(2R)$

$W_g = -(0.500)(9.8)(2\times 1)$

$W_g = -9.8 J$

Now by work energy equation we have

$\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2$

$\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2$

$v_f = 4.1 m/s$

Part c)

now minimum speed required at the top is such that the normal force must be zero

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = 3.13 m/s$

so here we got speed more than the required speed so it will reach the top

5 0

3 years ago

What amount of charge passes through a 3.0 amp television in 1.3 hours?

pogonyaev

Answer:

luv you

Explanation:

7 0

3 years ago

You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler

scZoUnD [109]

Answer:

q=1.4*10^{-9}C

Explanation:

Given data:

charge on ruler = -14μC

Mass of tissue is 5 g

To Know the minimum charge, equate electrostatic force to weight

we have F = W

so $\frac{KQq}{r^2} =mg$

putting all value in equation,

$=\frac{9*10^9*(14*10^{-6})*q}{0.06^2} = 5* 10^{-3}*9.8$

solving for q

$q =\frac{5* 10^{-3}*9.8 *0.06^2}{9*10^9*(14*10^{-6})}$

or q=1.4*10^{-9}C

5 0

3 years ago