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3 years ago

12

astone is thrown upwards with an initial velocity of 25m/s at ange of 30 to the ground. work out the stones horizantal range?

1 answer:

Vadim26 [7]3 years ago

4 0

Explanation:

$R = \dfrac{v_0^2}{g}\sin{2\theta} = \dfrac{(25\:\text{m/s})^2}{9.8\:\text{m/s}^2}\sin{60°}$

$\:\:\:\:=55.2\:\text{m}$

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Please, someone help me I'm begging yall 1. it's estimated that 1 kg of body fat will provide 3.8 * 10^7 J of energy. A 67 kg mo

Leto [7]

Answer:

0.24 kg used up

Explanation:

He has a mass of 67 kg

The gravitational constant is 9.81 m/s^2

The distance upward is 3500 m

W = m*g*h

W = 67 * 9.81 * 3500

Work = 2,300,445 Joules

Work = 2300 kj

work = 2.30 * 10^6 joules in scientific notation.

Part B

He needs 4 times this amount to climb the mountain because the body is only 25% efficient in converting energy.

4*2.30 * 10^6 = 9.20 * 10^6 Joules of energy are therefore required.

The total amount in a kg of fat = 3.8 * 10^7 joules

x kg of fat is needed to provide 9.20.*10^6 joules

1 kg / (3.8 * 10^7 J ) = x kg / (9.20 * 10^6 J)

9.20 * 10 ^6 * 1 = 3.8 * 10^7 *x

9.20 * 10 ^6 / 3.8 * 10^7 = x

x = 0.24 kg of fat are needed

7 0

3 years ago

A 480N sphere 40.0cm in radius rolls without slipping 1200cm down a ramp that is inclined at 53 0 with the horizontal. What is t

IgorC [24]

As we know that sphere roll without slipping so there is no loss of energy in this case

so here we can say that total energy is conserved

Initial Kinetic energy + initial potential energy = final kinetic energy + final potential energy

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2+ mgh'$

as we know that ball start from rest

$v_i = 0$

height of the ball initially is given as

$h = Lsin\theta$

$h = 1200sin53 = 960 cm$

also we know that

$I = \frac{2}{5}mR^2$

also for pure rolling

$v = r\omega$

also we know that

$480 = m*9.8$

$m = 49 kg$

now plug in all data in above equation

$480*9.60 + 0 = \frac{1}{2}*49*(0.40*\omega)^2 + \frac{1}{2}*\frac{2}{5}*49*(0.40)^2\omega^2 + 0$

$4608 = 3.92\omega^2 + 1.568\omega^2$

$\omega^2 = 839.65$

$\omega = 29 rad/s$

So speed at the bottom of the inclined plane will be 29 rad/s

7 0

4 years ago

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10

Mars2501 [29]

Answer:

a) $a=5.7551 \times g$

b) $d=20.5539\times g$

Explanation:

Given:

speed of rocket initially, $v_i=0\ m.s^{-1}$

top speed of rocket after acceleration, $v=282\ m.s^{-1}$

time taken to get to the top speed, $t_i=5\ m.s^{-1}$

final speed of the rocket, $v_f=0\ m.s^{-1}$

time taken to get to the final speed after reaching the top speed, $t_f=1.4\ s$

Now the acceleration:

$a=\frac{v-v_i}{t_i}$

$a=\frac{282-0}{5}$

$a=56.4\ m.s^{-2}$

Now as a fraction of gravity:

$a=\frac{56.4}{9.8}\times g$

$a=5.7551 \times g$

Now, the deceleration:

$d=\frac{0-282}{1.4}$

$d=201.4285\ m.s^{-2}$

Now as a fraction of gravity:

$d=\frac{201.4285}{9.8}\times g$

$d=20.5539\times g$

6 0

4 years ago

A rock falls off a cliff. How fast is the rock traveling vertically two seconds later?

Vikentia [17]

The rock it traveling really, really fast.
It is hard to exactly determine how fast bc u need the height of the cliff and how big the rock is.
Hope this helps and can I get brainliest answer!

8 0

3 years ago

Which of the following positions would Earth be in if it were winter in the Northern Hemisphere?

abruzzese [7]

Answer:

Earth's tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun's most direct rays. So, when the North Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it's winter in the Northern Hemisphere.

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3 years ago