Answer:
Shown by explanation;
Explanation:
The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)
Assumption;I assume the mass of the samples are : 109g and 192g
∆T= 30.1-21=8.9°c.
The heat of the samples are for 109g are:
0.109 × 4186 × 8.9 =4060.84J
For 0.192g are;
∆T= 67-30.1-=36.9°c
0.192 × 4186×36.9=29656.97J
Answer:
94.67 N
Explanation:
Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.
At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.

Substituting 70 N for W, 41 N for F and
for 37 degrees
N=70+41sin37=94.67441595 N and rounding off to 2 decimal places
N=94.67 N
Answer:
Approximately
, assuming friction between the vehicle and the ground is negligible.
Explanation:
Let
denote the mass of the vehicle. Let
denote the initial velocity of the vehicle. Let
denote the spring constant (needs to be found.) Let
denote the maximum displacement of the spring.
Convert velocity of the vehicle to standard units (meters per second):
.
Initial kinetic energy (
) of the vehicle:
.
When the vehicle is brought to a rest, the elastic potential energy (
) stored in the spring would be:
.
By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial
of the vehicle should be equal to the
of the vehicle. In other words:
.
Rearrange this equation to find an expression for
, the spring constant:
.
Substitute in the given values
,
, and
:

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