Answer:
Answer:
Bus travels 160 km in 4 hours
Speed of bus = 160/4 = 40 km/hr
Train travels 320 km in 5 hours
Speed of train = 320/5 = 64 km/hr
In one hour, bus travels 40 km and train travels 64 km.
Ratio = 40:64 = 5:8
Answer:
Moment of the force is 20 N-m.
Explanation:
Given:
Force exerted by the person is, 
Distance of application of force from the point about which moment is needed is, 
Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.
Therefore, the moment of the force about the end of the claw hammer is given as:

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.
I'm not sure but I know u is 10^6
Answer:
, the minus meaning west.
Explanation:
We know that linear momentum must be conserved, so it will be the same before (
) and after (
) the explosion. We will take the east direction as positive.
Before the explosion we have
.
After the explosion we have pieces 1 and 2, so
.
These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.
Since we know momentum must be conserved we have:

Which means (since we want
and
):

So for our values we have:

Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m