All categories

3 years ago

A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s,

1 answer:

3 0

By using the equation speed = distance/time we can solve for distance. The speed is 4 m/s and the time is 12 seconds. We need to rearrange the equation to Speed * Time = distance. 4(12) = 48; 48 = distance. The cliff is 48 meters high.

Read more about force here:

brainly.com/question/15300777

#SPJ1

6 0

2 years ago

How many minutes will it take a car to go from a stop to 33 km/hr if it accelerates at 10 km/hr²

Ber [7]

Answer:

33= 0+10 * t so t = 33 / 10 = 3.3 hrs

8 0

2 years ago

Read 2 more answers

A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

matrenka [14]

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

5 0

3 years ago

Which of the following is not a petroleum product

serg [7]

D. wool
hope it helped

4 0

3 years ago

ASAP please!! <br>The force vectors on an aircraft are as shown. Find the net (resultant force).

ioda

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

$W_D=3600 N\cdot\sin 27^\circ=1634N$

acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

The magnitude of the net force is:

$|F|=\sqrt{5366^2+829^2}N = 5430N$

6 0

3 years ago