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Masja [62]
3 years ago
6

What is the most problem of any defective tools?​

Physics
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer:

knife or weapon

Explanation:

that could be dangerous...

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An echo is not heard in a small room.why ? give reason<br>​
JulsSmile [24]

Answer:

They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the room should be 17.2 m.

hopw it helps

7 0
3 years ago
Read 2 more answers
Why is this event important to include in the biography?
nadya68 [22]

Answer:

D: It shows that Frida Kahlo used art to cope with her pain.

Explanation:

Within the text given it shows her emotions being lonely, immobile and in pain. But it all shows her asking her father for art which states that art is her sort of relief and happy place.

8 0
2 years ago
Read 2 more answers
What is the force of a ball if it has a mass of 10 kg and accelerates at an acceleration of 20 m/s^2? *
vampirchik [111]

Answer:

200N

Explanation:

mass(m) = 10 kg

acceleration(a) = 20 m/s^2

Force = mass * acceleration

          = 10*20

          = 200 N

Force = 200N

4 0
2 years ago
What are the possible units for a spring constant
AleksAgata [21]
Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)
5 0
3 years ago
Read 2 more answers
A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
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