Answer:
a) 400.4Joules
b) 262.69Joules
Explanation:
Work is said to be done if the force applied to an object cause the object to move through a distance
Workdone = Force × Distance
Given
Force = 76N
Distance= 5.2m
Work done = 77 × 5.2
Work done = 400.4Joules
b) If the force is exerted at an angle of 41°
Work done = Fdsin theta
Work done = 77(5.2)sin41
Work done = 400.4sin41
Work done = 262.69Joules
Answer:
(B) 0.5 g
Explanation:
Newton's second law says ∑ F i = m a .
the rate of change in momentum of a body is proportional to the force applied on the body.
f∝ma
f=kma
were k is constant and equal to 1
The centripetal acceleration is an acceleration.
the tension on the swing and object weight goes to the left hand side while the centripetal acceleration goes to the right handside
At the bottom of the swing, ΣF = FT – mg = mac;
notice that the tension in the swing is 1.5 times the weight of the object
we can write
1.5mg – mg = mac,
0.5mg = mac
0.5 g=ac
Answer:
Stage one of sleep, also known as the transitional phase, occurs when one finds themselves floating in and out of consciousness. During this NREM stage, you may be partially awake while your mind begins to drift off. This period of drowsiness eventually leads to a light sleep
Explanation:
i found it on google
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
a). Perihelion . . . the point in Earth's orbit that's closest to the Sun.
We pass it every year early in January.
b). Aphelion . . . the point in Earth's orbit that's farthest from the Sun.
We pass it every year early in July.
c). Proxihelion . . . a made-up, meaningless word
d). Equinox . . . the points on the map of the stars where the Sun
appears to be on March 21 and September 21.
