Answer:
The box displacement after 6 seconds is 66 meters.
Explanation:
Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (
), in meters, can be determined by the following expression:
(1)
Where:
- Initial velocity, in meters per second.
- Time, in seconds.
- Acceleration, in meters per square second.
If we know that 
, 
and 
, then the box displacement after 6 seconds is:
The box displacement after 6 seconds is 66 meters.
Savanna regions developed during the Triassic period. is true
Answer:
Explanation:
Given that,
Mass of sledge hammer;
Mh =2.26 kg
Hammer speed;
Vh = 64.4 m/s
The expression fot the kinetic energy of the hammer is,
K.E(hammer) = ½Mh•Vh²
K.E(hammer) = ½ × 2.26 × 64.4²
K.E ( hammer) = 4686.52 J
If one forth of the kinetic energy is converted into internal energy, then
ΔU = ¼ × K.E(hammer)
∆U = ¼ × 4686.52
∆U = 1171.63 J
Thus, the increase in total internal energy will be 1171.63 J.
Answer:
a) please find the attachment
(b) 3.65 m/s^2
c) 2.5 kg
d) 0.617 W
T<weight of the hanging block
Explanation:
a) please find the attachment
(b) Let +x be to the right and +y be upward.
The magnitude of acceleration is the same for the two blocks.
In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:
∑Fx=ma_x
T=m1a_x
14.7=4.10a_x
a_x= 3.65 m/s^2
c) <em>in order to calculate m we will apply newton second law on the hanging </em>
<em> block</em>
<em> </em>∑F=ma_y
T-W= -ma_y
T-mg= -ma_y
T=mg-ma_y
T=m(g-a_y)
a_x=a_y
14.7=m(9.8-3.65)
m = 2.5 kg
<em>the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax</em>
d) calculate the weight of the hanging block :
W=mg
W=2.5*9.8
=25 N
T=14.7/25
=0.617 W
T<weight of the hanging block
Answer:
40 ms¯².
Explanation:
To solve this problem, we shall illustrate the question with a diagram.
The attached photo gives a better understanding of the question.
From the attached photo:
Velocity (v) = 160 ms¯¹
Time (t) = 4 secs.
Acceleration (a) =?
Acceleration (a) = Velocity (v) /time (t)
a = v/t
a = 160/4
a = 40 ms¯²
Therefore, the initial acceleration of the rocket is 40 ms¯².
