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alexandr402 [8]

3 years ago

14

A ball is thrown downward from the top of a 55.0 m tower with an initial speed of 11.0 m/s. Assuming negligible air resistance,

what is the speed of the ball just before hitting the ground?

1 answer:

irina1246 [14]3 years ago

8 0

I'm not positive however my guess is this.
KE=(1/2)mv^2
KE=(1/2)(55)(121)
KE=3327.5m/s^2
You might not want to use this tho.

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A bus is moving and has 500000 joules of kinetic energy. The brakes are applied and the bus stops. How much work is needed to st

Bad White [126]

For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:
$W=K_f - K_i$
where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus

Since the bus comes at rest, its final kinetic energy is zero: $K_f = 0$ , so the work done by the brakes to stop the bus is
$W=-K_i = -500000 J$
And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.

8 0

3 years ago

A 20 ohm resistor has 210 volts measured<br> across it. What is the current?

AlexFokin [52]

Answer:

Given =

Resistance = 20 ohm

Volt = 210 volts

Solution =

By Ohm's Law

R = V/I

I = V/R

I = 210/20

I = 11.5 ampere. Answer

3 0

2 years ago

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago

Solved, can someone check it over!

kakasveta [241]

Answer:

its good no need to change anything :))

4 0

3 years ago

A runner runs around the track consisting of two parallel lines 96 m long connected at the ends by two semi circles with a radiu

Zielflug [23.3K]

-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero

7 0

3 years ago