Answer:
We can use 2 g H = v2^2 - v1^2 or
v2^2 = 2 g H + v1^2
Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec
The object will return with the same speed that it had initially so the object
starts out with a downward speed of 44 ft/sec
Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2
v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2
v2 = 110 ft/sec
Answer: Option B. R = (1/2)gt^2
Explanation:
S = R (horizontal distance)
V^2 = 2gS
V^2 = 2gR
R = V^2 / 2g
But V = gt
R = (gt)^2 / 2g
R = (g^2 x t^2) / 2g
R = gt^2 / 2
But t^2 = 2h/g
R = ( g x 2h/g) / 2
R = h
But h = (1/2)gt^2
R = h = (1/2)gt^2
<h3>
Answer:</h3>
1.5 m/s²
<h3>
Explanation:</h3>
We are given;
Force as 60 N
Mass of the Cart as 40 kg
We are required to calculate the acceleration of the cart.
- From the newton's second law of motion, the rate of change in momentum is directly proportional to the resultant force.
- That is, F = ma , where m is the mass and a is the acceleration
Rearranging the formula we can calculate acceleration, a
a = F ÷ m
= 60 N ÷ 40 kg
= 1.5 m/s²
Therefore, the acceleration of the cart is 1.5 m/s²
Answer:
W = 0.060 J
v_2 = 0.18 m/s
Explanation:
solution:
for the spring:
W = 1/2*k*x_1^2 - 1/2*k*x_2^2
x_1 = -0.025 m and x_2 = 0
W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2
W = 0.060 J
the work-energy theorem,
W_tot = K_2 - K_1 = ΔK
with K = 1/2*m*v^2
v_2 = √2*W/m
v_2 = 0.18 m/s
Answer:
stryo: 1
wood: 1
ice: 1
brick: 2
aluminum: 2.7
Explanation:
d= mass/ total volume
(fyi: for aluminum, they did the subtraction wrong to find the total volume. it is actually 5 or 5.00)
