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AlladinOne [14]
2 years ago
5

A rock falls off a cliff with an acceleration of -9.8 m/s^2 and hits the ground 5 s later . How high is the diff?​

Physics
1 answer:
lara [203]2 years ago
6 0
50 m/s is the answer
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A pencil is rolled off a table of height 0.92 m. If it has horizontal speed pf 1.4 m/s, how long does it take the pencil to reac
Alenkasestr [34]

The distance an object falls, from rest, in gravity is

                         D  =  (1/2) (G) (T²)

                        'T' is the number seconds it falls.

In this problem,

                         0.92 meter = (1/2) (9.8) (T²)

Divide each side by  4.9 :   0.92 / 4.9 = T²

Take the square root
of each side:                          √(0.92/4.9) = T

                                                  0.433 sec = T    

The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor.  BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.

The pencil is in the air for  0.433 second.
In that time, it travels
                                   (0.433s) x (1.4 m/s) = 0.606 meter

from the edge of the table.
 
3 0
3 years ago
A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w
katovenus [111]

Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85  = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x  5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

= .38788  x 10⁻³ C .

= 387.88 x 10⁻⁶ C .

4 0
3 years ago
The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated i
gregori [183]

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

I = 4.2 kg × 0.0626 m²

I = 0.2625 kg.m²

So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

∝ = T / I

we substitute

∝ = 4.104 / 0.2625

∝ = 15.65 rad/s²

Therefore, the resulting angular acceleration is 15.65 rad/s²

8 0
3 years ago
Please help on this one?
diamong [38]

Answer:

ITS C

Explanation:

7 0
2 years ago
How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
solong [7]

Answer:

71 rpm

Explanation:

Given that:

Angular momentum (L) = 0.26

Diameter = 25cm = 0.25 cm

Radius, r = (d/2) = 0.125m

Mass = 5.6 kg

Moment of inertia (I) = 2mr² / 5

I = (2 * 5.6 * 0.125^2) / 5

= 0.175

= 0.175 / 5

= 0.035 kgm²

Angular speed (w) ;

w = L / I

w = 0.26 / 0.035

= 7.4285714

= 7.429 rad/s

w = (7.429 * 60/2π)

w = 445.74 / 2π rpm

w = 70.941724

Angular speed = 70.94 rpm

= 71 rpm

5 0
2 years ago
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