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Bas_tet [7]
3 years ago
14

Which describes how the same force affects a small mass and a large mass

Physics
1 answer:
n200080 [17]3 years ago
6 0
The same force accelerates a small mass faster than
it accelerates a large mass.

It's easier to get a little red wagon going by pushing it
than it is to get a school bus going by pushing it.
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Which of the following are not properties of metals? Check all that apply. *
wolverine [178]

D. they have a full luster

8 0
3 years ago
- The school zone in front of your school has a posted speed limit of 25 mi/h, which is about 11 m/s. Let's
miv72 [106K]

Answer:

s = 38.7 m

Explanation:

First we calculate the distance covered during uniform motion of reaction time.

s₁ = vt

where,

s₁ = distance covered during uniform motion = ?

v = uniform speed = 11 m/s

t = time = 2.3 s

Therefore,

s₁ = (11 m/s)(2.3 s)

s₁ = 25.3 m

Now, we calculate the distance covered during decelerated motion:

2as₂ = Vf² - Vi²

where,

a = deceleration = -4.5 m/s²

s₂ = distance covered during decelerated motion = ?

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 11 m/s

Therefore,

2(-4.5 m/s²)s₂ = (0 m/s)² - (11 m/s)²

s₂ = (-121 m²/s²)/(-9 m/s²)

s₂ = 13.4 m

the total distance will be:

s = s₁ + s₂

s = 25.3 m + 13.4 m

<u>s = 38.7 m</u>

5 0
3 years ago
For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant
Scilla [17]
Consider the projectile launched at initial velocity V at angle θ relative to the horizontal.
Neglect wind or aerodynamic resistance.

The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g =  acceleration due to gravity.

The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
    = (2V²sinθ cosθ)/g
    = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, \frac{dD}{d \theta}  =0
That is,
\frac{2V^{2}}{g} cos(2 \theta )=0

Because \frac{2V^{2}}{g}  \neq 0, therefore cos(2θ) = 0.
This is true when 2θ = π/2  => θ = π/4.

It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.

4 0
4 years ago
What is the fundamental cause of air circulation in earths atmosphere
Digiron [165]
<span>A differences in the warmth, or moisture level as well as neighbouring areas of pressure in </span>air cause air<span> to circulate. in the earth's atmosphere.</span>
4 0
4 years ago
A point source of light is 1.24 m below the surface of a pool. What is the diameter of the circle of light that a person above t
Ksivusya [100]

Answer:

D = 2.828 m

Explanation:

given,

distance of source of light = 1.24 m below surface of pool

refractive index of the water  = n₁ = 1.33

refractive index of air = n₂ = 1

refraction angle be = 90°

let C be the critical angle

Radius = d tan C

d is the depth of the source

Using Snell's law

n₁ sin C = n₂ sin R

1.33 x sin C = 1 x  sin 90°

sin C = \dfrac{1}{1.33}

C = sin^{-1}(0.752)

C = 48.75°

hence,

R = 1.24 x tan 48.75°

R = 1.414 m

Diameter = 2 x R

D = 2 x 1.414

D = 2.828 m

5 0
3 years ago
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