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mart [117]
3 years ago
6

Space awareness is a concept that needs to be taught at the beginning of a physical education program because

Physics
1 answer:
castortr0y [4]3 years ago
3 0
Future lies in space program, as we humans have to explore outer planets and sapce so, space awareness is must
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An elevator ascends at a constant speed of 4 m/s, how much time is required for the elevator in order to travel 120 m upwards?
Dominik [7]

Divide 120 by 4, to get 30 seconds

5 0
3 years ago
Read 2 more answers
Help me please ty<br><br> .......................
Vikki [24]

Answer:

maybe c

Explanation:

but im not sure though

3 0
3 years ago
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A rifle fires a 2.05 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the
miss Akunina [59]

Answer:

Spring Constant = 279.58 N/m

Explanation:

We are given;

Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg

Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m

Maximum height; h = 4.46 m

The formula for the energy in the spring is given by;

E = ½kx²

where:

k is the spring constant

x is the distance the spring is compressed.

Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;

½kx² = mgh

Plugging in the relevant values, we have;

½ * k * 0.0801² = 0.0205 * 9.81 * 4.46

0.003208005k = 0.8969

k = 0.8969/0.003208005

k = 279.58 N/m

6 0
3 years ago
A wave is incident on the surface of a mirror at an angle of 41° with the normal. What can you say about its angle of reflection
bazaltina [42]
Hold on 63 deg angle my friend
7 0
4 years ago
Read 2 more answers
A 4.00-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 27.0 N is require
NikAS [45]

Answer:

a)

135Nm⁻¹

b)

0.925 Hz

c)

1.2ms⁻¹

d)

0 m

e)

6.7ms⁻²

f)

\pm 0.2 m

Explanation:

a)

F = force required to hold the object at rest connected with stretched spring = 27 N

x = stretch in the spring from equilibrium position = 0.2 m

k = force constant of the spring

force required to hold the object at rest is same as the spring force , hence

F = k x

k = \frac{F}{x}

inserting the values

k = \frac{27}{0.2} = 135 Nm⁻¹

b)

frequency of the oscillations is given as

f =\frac{1}{2\pi }\sqrt{\frac{k}{m}}

inserting the values

f =\frac{1}{2(3.14) }\sqrt{\frac{135}{4}}\\f = 0.925 Hz

c)

A = Amplitude of oscillations = 0.2 m

w = angular frequency

Angular frequency is given as

w = 2\pi f = 2 (3.14) (0.925) = 5.8 rads⁻¹

Maximum speed of oscillation is given as

v_{max} = Aw

v_{max} = (0.2)(5.8)\\v_{max} = 1.2 ms⁻¹

d)

maximum speed of the object occurs at the equilibrium position, hence

x = 0 m

e)

Maximum acceleration of oscillation is given as

a_{max} = Aw^{2}

a_{max} = (0.2)(5.8)^{2}\\a_{max} = 6.7ms⁻²

f)

maximum acceleration occurs when the object is at extreme positions, hence

x = \pm 0.2 m

4 0
3 years ago
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