Answer
given,
force = 94 lb
weight of crate = 220 lb
Assuming the static friction be equal = 0.47
kinetic friction = 0.36
Maximum force applied to move the object is when object is just start to move.
F = μ N
F = 0.47 x 220
F = 103.4 lb
As the frictional force is more than applied then the object will not move.
so, the friction force will be equal to the force applied on the object that is equal to 94 lb.
hence, the direction of force will left.
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Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Answer:
13.4 x 10 raise to power -19 C
Explanation:
. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m
. The uniform electric field is E = 214 N/M
, The decrease in electrical potential energy is
d(P.E) = 51.63 x 10 raise to power -19 J
Let the magnitude of the charge of the moving particle be q
which is given by the equation
d(P.E) =qEd
51.63 x 10 power -19 = q(214)(0.018)
51.63 x 10 power -19 =3.852q
by making q the formular,
q = 13.4 x 10 power -19 C
When somebody hands you a Celsius°, it's easy to find the equivalent Fahrenheit°.
Fahrenheit° = (1.8 · Celsius°) + 32° .
So 100°C works out to 212°F.
It's also easy to find the equivalent Kelvin. Just add 273.15 to the Celsius.
So now you can see that 100°C is equal to A and D,
and it's less than B .
The only one it's greater than is C .