Answer:
a)52.58 m/s
b)56.13°
Explanation:
assume the upward direction as positive
x-component of the velocity = 29.3×cos33.6°=24.40 m/s (remain constant)
y-component of the velocity which is -29.3sin33.6°= -16.21 m/s
time of flight = 68.3/24.40= 2.7991 seconds
now, we can obtain final velocity in y-direction


=43.66 m/s

=52.58 m/s
for direction

56.13° from the horizontal
Answer:
F = M a is the vector equation involved
F = T - M g are the forces acting on the elevator (scalar equation)
T - M g = M a
T = M (a + g) remember this a scalar
If a is slowing down then it must have a positive acceleration upwards
Therefor the tension in the cable must be greater than zero
When the tension increases to M g, a has increased to zero
For a to be zero, no acceleration, T = M g
Answer:
(a) 11.66 square meters per liter
(b) 11657.8 per meters
(c) 0.00211 gal per square feet
Explanation:
(a) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 1gal/3.7854L = 11.66m^2/L
(b) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 264.17gal/1m^3 = 11657.8/m
(c) Inverse of 475ft^2/gal = 1/475ft^3/gal = 0.00211gal/ft^3
Answer:
9) This is a case of deceleration
10)-0.8 ms-2
b) acceleration is the change in velocity with time
11)
a) 100 ms-1
b) 100 seconds
12) 10ms-1
13) more information is needed to answer the question
14) - 0.4 ms^-2
15) 0.8 ms^-2
Explanation:
The deceleration is;
v-u/t
v= final velocity
u= initial velocity
t= time taken
20-60/50 =- 40/50= -0.8 ms-2
11)
Since it starts from rest, u=0 hence
v= u + at
v= 10 ×10
v= 100 ms-1
b)
v= u + at but u=0
1000 = 10 t
t= 1000/10
t= 100 seconds
12) since the sprinter must have started from rest, u= 0
v= u + at
v= 5 × 2
v= 10ms-1
14)
v- u/t
10 - 20/ 25
10/25
=- 0.4 ms^-2
15)
a=v-u/t
From rest, u=0
8 - 0/10
a= 8/10
a= 0.8 ms^-2
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