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3 years ago

The purple arrow shows speed and direction of the ball. This arrow is called what?

Physics

2 answers:

valina [46]3 years ago

6 0

Vector I'm pretty certain

Lapatulllka [165]3 years ago

5 0

The arrow is called vector

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Good day can I get some help please?

abruzzese [7]

Answer:

432 J

Explanation:

When moving linearly:

Kinetic Energy = (1/2)mV^2

So here you have:

KE=(1/2)(6)(12^2)=(1/2)(6)(144)=432

The unit for energy is Joules (J), so your answer would be 432 J.

8 0

4 years ago

Could someone help me with this please?

ycow [4]

Answer:

you could measure several properties of

the unknown liquid and compare them with the properties of known

substances. You might observe and measure such properties as color,

odor, texture, density, boiling point, and freezing point.

6 0

3 years ago

Read 2 more answers

Fredrick did an investigation on mass and acceleration. He applied a force to a toy car and measured the distance it moved. He t

garik1379 [7]

Explanation:

B. More mass results in less acceleration.

6 0

3 years ago

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

3 0

3 years ago

How high would you have to lift a 1000kg car to give it a potential energy of:

Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

It is mathematically expressed as:

P.E = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

h = $\frac{P.E }{mg}$

For 2.0 x 10³ J;

h = $\frac{2000}{1000 x 9.8}$ = 0.204m

For 2.0 x 10⁵ J;

h = $\frac{200000}{9.8 x 1000}$ = 20.4m

For 1.0kJ = 1 x 10³J;

h = $\frac{1000}{9.8 x 1000}$ = 0.102m

5 0

3 years ago