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fgiga [73]
3 years ago
7

A supersonic jet is at an altitude of 14 kilometers flying at 1,500 kilometers per hour toward the east. At this velocity, how f

ar will the jet fly in 1.6 hours?
Physics
1 answer:
Kaylis [27]3 years ago
3 0

Answer:

The jet will fly 2400 km.

Explanation:

Given the velocity of the jet flying toward the east is 1,500 kmph toward the east.

We need to find the distance covered in 1.6 hours.

In our problem we are given speed and time, we can easily determine the distance using the following formula.

Distance=Speed\times Time

Distance=1500\times 1.6=2400\ km

So, the supersonic jet will travel 2400 km in 1.6 hours toward the east from its starting point.

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If you place a balloon in the freezer, it will __________.
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if a ball is rolling at a velocity of 1.5 m/sec and has the momentum of 10.0 kg m/sec, what is the mass of the ball?
PolarNik [594]
Okay. There is a simple formula to go with this where:

p = mv

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M: Mass.
V: Velocity

Sub the numbers in and solve for M.

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6 0
3 years ago
A spherical balloon is being inflated and the radius of theballoon is increasing at a rate of 2 cm/s.(A) Express the radius (r)
goblinko [34]

Answer:

A.) r = 2t

B.) V = 33.5t^3

Explanation:

Given that a spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2 cm/s

A) Express the radius (r) of the balloon as a function of the time (t).

Since the rate = 2 cm/s that is,

Rate = radius/ time

Therefore,

2 = r/t

Make r the subject of formula

r = 2t

(B) If V is the volume of the balloon as a function of the radius, find V or and interpret it.

Let assume that the balloon is spherical. Volume of a sphere is;

V = 4/3πr^3

Substitute r = 2t into the formula

V = 4/3π(2t)^3

V = 4/3π × 8t^3

V = 32/3 × πt^3

V = 33.5t^3

6 0
3 years ago
A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to
SpyIntel [72]
This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

 46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s


3 0
3 years ago
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