All categories

3 years ago

When naming a ray does the direction of the arrow matter?

Physics

1 answer:

mixas84 [53]3 years ago

7 0

The first letter of the ray is the point where the ray starts at.
The second letter is where the arrow is going.

You might be interested in

Atomic math challenge will give brainly and thanks

nevsk [136]

Answer:

1. Hydrogen

Atomic # = 1

Atomic Mass = 1.00794 ( If you round it it's 1.008 )

# of protons = 1

# of neutrons = none

# of electrons = 1

8 0

3 years ago

Read 2 more answers

A pair of fuzzy dice is hanging by a string from your rearview mirror. while you are accelerating from a stoplight to 28 m/s in

Lerok [7]

We can use the formula of motion in physics (2nd law od newton) in this problem:
x direction: Fsin ∅ = ma
y direction: Fcos ∅ -mg = 0
∅ is equal to sin ∅ / cos ∅ or x/y
tan ∅ = ma / mg = a /g

Applying acceleration formula:
v = vo + at ; 28 = 0 + 6a ; a = 4.67 m/s^2
∅ = tan-1 (a/g) = tan-1 (4.67/9.81) = <span>25.4 degrees.</span>

4 0

3 years ago

based on how the book describes a heat pump in sec 16.3 how does a heat pump prove law of conservation of energy

sattari [20]

The law of conservation of energy states that: "energy cannot be created nor destroyed. It can only be transformed from one form to the other". A heat pump transfers heat from a colder area to a warmer area. As the conservation of energy goes, the energy that comes in, must equal the energy that comes out. Although no machine is 100% efficient, the missing energy is supplied by an external energy source, such as electricity.

4 0

4 years ago

2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or

krek1111 [17]

a) The work done by the tree is $-1.44\cdot 10^5 J$

b) The amount of force applied is 2880 N

Explanation:

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

$W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

$W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J$

The work is negative because it is done in the direction opposite to the direction of motion of the car.

The work done by the tree on the car can also be rewritten as

$W=Fd$

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0

4 years ago

¡¡first convert revolutions per second to

katovenus [111]

Answer:

a. 5.23 m/s² b. 44.23 N

Explanation:

a. What is the centripetal acceleration of the hammer?

The centripetal acceleration a = rω² where r = radius of circle and ω = angular speed.

Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.

So a = rω²

= 1.4 m × (3.74 rad/s)²

= 5.23 m/s²

b. What is the tension in the chain?

The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.

So, T = 8.45 kg × 5.23 m/s²

= 44.23 N

6 0

3 years ago