Examine the weather map and locate this symbol Air Weather Climate part 1

5. You are skateboarding on top of a small hill and take a break. You have 100 J of potential energy. If

Sholpan [36]

Given the potential energy of 100 J, the height of the hill is 0.128 m (12.8 cm)

Explanation:

The gravitational potential energy of an object is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object, relative to the ground

In this problem, we have

U = 100 J is the potential energy of the person

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

m = 80 kg is the mass of the person

Solving for h, we find at what height the person is:

$h=\frac{U}{mg}=\frac{100}{(80)(9.8)}=0.128 m = 12.8 cm$

Learn more about gravitational potential energy here:

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5 0

3 years ago

ARRANGE THE STEPS IN ORDER TO DESCRIBE WHAT HAPPENS TO A GAS WHEN IT COOLS

vredina [299]

What happens to has when it cools is ...
Step 1) They will start to form liquids, like condensation.
Step 2) As liquids cool, they will turn into solids.
Step 3) As solids cool, they become more stable and solid.

3 0

3 years ago

The electric potential in a certain region varies with position as V(x) = ax2 - bx + c, where a = 2.2 V/m2, b = 16.5 V/m, and c

Effectus [21]

Answer:

idk it its A

Explanation:

i did it

6 0

3 years ago

Two cars are traveling at the same speed of 27 m/s on a curve thathas a radius of 120 m. Car A has a mass of 1100 kg and car B h

Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

$a_{cB} = 6.075$ m/s²

$F_{cA} = 6682.5 N$

$F_{cB} = 9720 N$

Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

∑F = m*a Formula ( 2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

Data

$v_{A} = 27 \frac{m}{s}$

$v_{B} = 27 \frac{m}{s}$

$m_{A} = 1100 kg$

$m_{B} = 1600 kg$

r= 120 m

Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

4 0

4 years ago

In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let’s find out by doing a

mixer [17]

Answer:

0.07142 m/s in the opposite direction of the bullet

Explanation:

$m_1$ = Mass of rifle = 70 kg

$m_2$ = Mass of bullet = 0.01 kg

$v_1$ = Velocity of rifle

$v_2$ = Velocity of bullet = 500 m/s

As the momentum of the system is conserved

$m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.01\times 500}{70}\\\Rightarrow v_1=-0.07142\ m/s$

The recoil velocity of the rifle is -0.07142 m/s

The negative sign shows the opposite direction of the rifle.

3 0

3 years ago