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Greeley [361]
3 years ago
13

Difference between rest and motion?

Physics
1 answer:
Yuki888 [10]3 years ago
6 0

Answer:

Rest and motion are relative terms. In simple terms, an object that changes its position is said to be in motion while the opposite action causes an object to be at rest.

Explanation:

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What would you use to to measure the weight of a truck?? I’m needing a word that is 8 letters long
myrzilka [38]

Answer:

Railroad Scale?

Explanation:

5 0
3 years ago
an object with a mass of 2.4 kg has a force of 12.6 N applied to it. What is the resulting acceleration of the object? WITH PROO
Scorpion4ik [409]
Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2
3 0
3 years ago
An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If  Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass m_1 = 30 kg

mass m_2 = 50 kg

\mu = 0.1

From the question; we can arrive at two cases;

That :

m_{2} a _ \ {net} }= m_2g - T   ----- equation (1)

m_{1} a _ \ {net} }=  T - mg sin \theta  - F ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a =[ 50  - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]g

80 a = 32. 4 × 10 m/s ²  (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0
3 years ago
The standard metric unit of density for a solid is
prisoha [69]
There is no SI "base unit" of density.
(Any unit of mass) divided by (any unit of volume) is
a valid unit of density.

The units of density that are seen most often are

(gram per cm³) and (kgm per meter³) .
5 0
3 years ago
We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

R = 2(1.5 -1)\ 30

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0
3 years ago
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