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Leokris [45]
3 years ago
9

The type of lens that spreads out parallel light is a

Physics
1 answer:
ohaa [14]3 years ago
7 0

the answer is concave lens

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A 85-kg astronaut is stranded from his space shuttle. He throws a 1-kg hammer away from the shuttle with a velocity of 17 m/s. H
algol [13]

Answer:

0.2 m/s

Explanation:

given,

mass of astronaut, M = 85 Kg

mass of hammer, m = 1 Kg

velocity of hammer , v =17 m/s

speed of astronaut, v' = ?

initial speed of the astronaut and the hammer be equal to zero = ?

Using conservation of momentum

(M + m) V = M v' + m v

(M + m) x 0 = 85 x v' + 1 x 17

85 v' = -17

  v' = -0.2 m/s

negative sign represent the astronaut is moving in opposite direction of hammer.

Hence, the speed of the astronaut is equal to 0.2 m/s

4 0
3 years ago
Where are a hurricanes fastest winds and heaviest rain found?
PSYCHO15rus [73]
The fastest winds and heaviest rain is found in the eye of the hurricane.
8 0
3 years ago
A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At
Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

6 0
2 years ago
A student on her way to school walks eastward in a straight line 20.0 meters towards the bus stop, but realizes she dropped her
larisa [96]

Answer:

Total displacement will be 47 meter

Total distance will be 83 meters

Explanation:

We have given that first the student go eastward towards bus stop 20 meters

But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters

So displacement = 20-18 = 2 meters

And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters

Total distance traveled by the student = 20+18+45 = 83 meters  

3 0
3 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0
3 years ago
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