Answer:
261.3 m/s
Explanation:
Mass of bullet=m=15 g=
1 kg=1000g
Mass of block=M=3 kg
d=0.086 m
Total mass =M+m=3+0.015=3.015 kg
K.E at the time strike=Gravitational potential energy at the end of swing

Using g=
Substitute the values




Velocity after collision=V=1.3 m/s
Velocity of block=v'=0
Using conservation law of momentum

Using the formula




Answer:
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
Explanation:
Given that
Yield strength ,Sy= 240 MPa
Tensile strength = 310 MPa
Elastic modulus ,E= 110 GPa
L=380 mm
ΔL = 1.9 mm
Lets find strain:
Case 1 :
Strain due to elongation (testing)
ε = ΔL/L
ε = 1.9/380
ε = 0.005
Case 2 :
Strain due to yielding


ε '=0.0021
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
For computation of load strain due to testing should be less than the strain due to yielding.
Answer:
250,000
Explanation:
<h2> </h2>
<h2>formula = ( F=ma </h2>
- F=1500N
- a=6m/s^2
- F= ma
- m=?
- 1500/6 = m
- m=250 kg
- 1kg =1000gm so 250kg =250,000gm
- m =250×10^3 gm
Answer:
No
Explanation:
The force of tension exerted by the string on the rock acts as centripetal force, so its direction is always towards the centre of the circle.
However, the direction of motion of the rock is always tangential to the circle: this means that the force is always perpendicular to the direction of motion of the rock.
As we know, the work done by a force on an object is

where
F is the magnitude of the force
d is the displacement of the object
is the angle between the force and the displacement
In this situation, F and d are perpendicular, so
, therefore
and the work done is zero:

<span>A van is traveling on a road at a speed of 55 km/h relative to a
stationary observer on the side of the road. A girl sitting near the
driver of the van throws a paper airplane to a boy at the back of the
van with a speed of 2 km/h relative to the girl, the boy, and the van.
The speed of the paper airplane, relative to the same stationary observer
on the side of the road, is (55 - 2) = 53 km/h. No rounding is necessary.</span>