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alexandr402 [8]
3 years ago
14

A variable that is described using both a number and direction is called

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

they are called a vector.

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2. A 50 kg diver is standing on the edge of a 15 m high cliff. What is his potential energy?
Lera25 [3.4K]

Answer:

3675 J

Explanation:

Gravitational Potential Energy = \frac{1}{2} × mass × g × height

( g is the gravitation field strength )

Mass = 50 kg

G = 9.8 N/kg ( this is always the same )

Height = 15 m

Gravitational Potential Energy = \frac{1}{2} × 50 ×9.8 × 15

= 3675 J

3 0
3 years ago
Which surface has the most friction? a An ice rink b A grassy field c A paved road
vovangra [49]

Answer:

An Ice Rink

Explanation:

7 0
3 years ago
Read 2 more answers
What are dimensionless quantities??​
bearhunter [10]

Answer:

Characteristic numbers are dimensionless numbers used in fluid dynamics to describe a character of the flow. To compare a real situation with a small-scale model it is necessary to keep the important characteristic numbers the same. Names of these numbers were standardized in ISO 31, part 12.

Explanation:

7 0
2 years ago
Which subatomic particle has the smallest mass?
spin [16.1K]

Answer:

quark

Explanation:

neutrón= 1.675 x 10-27 kg

proton=1,673 x 10-27 kg

electron=9,11 x 10-31 kg

quark=7,13 - 14,26 · 10−30 kg

3 0
2 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
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