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lara31 [8.8K]
3 years ago
12

A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is

Physics
1 answer:
Lostsunrise [7]3 years ago
4 0

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 ×  40 × a  = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

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Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
A roller-skate with a mass of 10kg has a force of 12N exerted on it. At what rate did the skate accelerate?
nataly862011 [7]

Answer:

1.2ms⁻²

Explanation:

F = ma

12 = 10*a

12/10 = a

6/5 ms⁻² = a   or 1.2 ms⁻² = a

6 0
3 years ago
Can anyone pls help me with this I’ll appreciate it. I’m trying to study for a test but I don’t know the answer to this.
forsale [732]

Answer:

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Explanation:

7 0
3 years ago
A cheetah can run at a maximum speed
julsineya [31]

Answer:

Approximately 17.21 seconds

Explanation:

With subtraction, we have the gazelle 18.3 km/h slower than the cheetah, which is about 5.08333 m/s. As the gazelle is 87.5 meters ahead of the cheetah, 87.5 divided by 5.083333333 is about 17.21 seconds.

5 0
3 years ago
You have four identical conducting spheres: A, B, C, and D. In each scenario below, sphere A starts with a charge of -Q, sphere
mash [69]

Answer:

1)    Q_c = - ½ Q ,  2)   Q_c = + ¼ Q , 3)    Q_c= 3/8 Q

Explanation:

For this exercise we must use that equal charges repel and charges with different signs attract. When two objects are in contact, the charges are evenly distributed between them.

Scenario 1.

when the two spheres touch the charge -Q is distributed between them, when separating each sphere has a charge = -1/2 Q

as there are no more interacts sphere C its charge is

         Q_c = - ½ Q

Scenario 2

when the two spheres touch the charge -Q is distributed between them, when separating each one has a charge

        Q_a = Q_d = - ½ Q

now the sphere D and C touch the charge is Q_net = -1/2 Q + Q = + ½ Q

when separating each sphere has half the charge

            Q_d = Q_c = + ¼ Q

since sphere C has no more interaction, its charge is

           Q_c = + ¼ Q

Scenario 3

A and B touch the net charge is Q_net = - Q + 0 = - Q

when parting

          Q_a = Q_b = - ½ Q

now B and D touch, the charge is Q_net = - ½ Q +0 = - ½ Q

when parting

           Q_b = Q_d = - ¼ Q

finally C and D touch

the net charge is Q_net = Q- ¼ Q = ¾ Q

when separating each one is left with half the load

           Q_c = Q_d = 3/8 Q

          Q_c= 3/8 Q

8 0
3 years ago
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