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Black_prince [1.1K]
3 years ago
10

A cyclist rode an average speed of 10 miles per hour for 15 miles how long was the ride

Physics
1 answer:
Ainat [17]3 years ago
7 0
I think it's an hour and a half
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A ball is thrown directly downward with an initial speed of 7.65 m/s froma height of 29.0 m. After what time interval does it st
coldgirl [10]

Answer: 1.77 s

Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:

xf= xo+vo*t+(g*t^2)/2  we can consider the origin on the top so the xo=0 and xf=29 m; then

(g*t^2)/2+vo*t-xf=0  vo is the initail velocity, vo=7.65 m/s

then by solving the quadratric equation in t

t=1.77 s

8 0
3 years ago
I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
Ad libitum [116K]

Answer:

d = 11.1 m

Explanation:

Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

\frac{1}{2} m {v}^{2}  = mgh

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

\sin(15)  =  \frac{h}{d}  \\ or \: h = d \sin(15)

Plugging this into the energy conservation equation and cancelling m, we get

{v}^{2}  = 2gd \sin(15)

Solving for d,

d =  \frac{ {v}^{2} }{2g \sin(15) }  =  \frac{ {(7.5 \:  \frac{m}{s}) }^{2} }{2(9.8 \:  \frac{m}{ {s}^{2} })(0.259)}   \\ = 11.1 \: m

3 0
3 years ago
Can someone PLEASE HELP ME!!!!!!!!
sweet [91]
<h3>WATER</h3>

Explanation:

<h2>#CARRYINGTOLEARN:)</h2>
7 0
3 years ago
Read 2 more answers
Greg is in a bike race. at mile marker four (out of ten), his speed was measured at 13.5 mph. which best describes the measured
posledela
The instantaneous speed. 
3 0
3 years ago
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A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point locat
stellarik [79]

Answer:

(a) 17.37 rad/s^2

(b) 12479

Explanation:

t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0

w = v / r = 99 / 0.06 = 1650 rad/s

(a) Use first equation of motion for rotational motion

w = w0 + α t

1650 = 0 + α x 95

α = 17.37 rad/s^2

(b) Let θ be the angular displacement

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 α θ

1650^2 = 0 + 2 x 17.37 x θ

θ = 78367.87 rad

number of revolutions, n = θ / 2 π

n = 78367.87 / ( 2 x 3.14)

n = 12478.9 ≈ 12479

4 0
3 years ago
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