The distance of the ship from the shore is 1020m
<u>Explanation:</u>
Given:
Speed, s = 340 m/s
Time, t = 10s
Distance, x = ?
The sound is going to have to go to shore, then come back.
The total round-trip distance is
D = speed X time
D = (340 m/s) * (6.0 s)
D = 2,040 m
But as previously stated, the sound had to get there, then come back. So the actual ship-to-shore distance is only half that.
x = D/2
Therefore, the distance of the ship from the shore is 1020m
Answer:
-(0.330m/s² ) kˆ
Explanation:
given data:
Mass of particle 'm'= 1.81 x kg
Velocity 'v'= (3.00 x m/s)j
Charge of particle 'q'= 1.22 x C
Uniform magnetic field 'B' = (1.63iˆ + 0.980jˆ )T
In order to calculate particle's acceleration, we'll use Newton's second law of motion i.e F=ma
Also,the force a magnetic field exerts on a charge q moving with velocity v is called the magnetic Lorentz force. It is given by:
F = qv × B
F= ma = qV x B
a= --->eq(1)
Lets determine the value of (v x B) first
v x B= (3.00 x m/s)j x (1.63iˆ + 0.980jˆ )
v x B= 4.89 x
Plugging all the required values in eq(1)
a= [1.22 x x (4.89 x kˆ)] / 1.81 x
a= -(0.330m/s² ) kˆ
-ve sign is representing the opposite direction
The answer is 17500 J for 35*10*50 and the answer is 16500 J for 55*10*30. Thus , the first object has more gravitational potential energy
The airplane's speed relative to the ground is
√ (100² + 25²)
= √ (10,000 + 625)
= √ 10,625
= 103.08 km/hr .
The angle of its velocity north of west is
the angle whose tangent is (25/100)
arctan(25/100) = 14° north of west .
(bearing = 284°)
Two thermometers, calibrated in celsius and fahrenheit respectively, are put into a liquid. the reading on the fahrenheit scale is four times the reading on the celsius scale. the temperature of the liquid is: