M-? P=800kg /m3 V=0,5L

Three diffrent examples of accelerated motion

LekaFEV [45]

Answer:

The three different examples of the accelerated motion are Falling/dropping of ball, Standing in circular rotating space, moving around the circle.

Explanation:

Acceleration is the change in velocity, which is related to the speed and direction in which the object is travelling. Hence, speeding up, slowing down and turning are few types . A simple example would be dropping a ball: as it falls its speed increases, which is a type of acceleration. A more complicated example would be standing in a circular, rotating space station. A point on the station moves in a circle, meaning that as it travels it must be turning (to remain in circular motion) making this another example of acceleration

3 0

3 years ago

What is the minimum force require to move a 5kg wooden crate on a wooden floor?

kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with µ. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ F = p - f = 0

• net vertical:

∑ F = n - w = 0

where

p = magnitude of the pushing force

f = mag. of friction

n = mag. of the normal force

w = weight of the crate

The second equation gives

n = w = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of µ, so

f = µ (49 N) = 49µ N

To overcome static friction, the push has to exceed this in magnitude, so that

p > 49µ N

For instance, if p = 0.25, then p would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

7 0

3 years ago

When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these

Zigmanuir [339]

Answer:

a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

v = $\sqrt{ \frac{T}{ \mu } }$

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

m = 2m₀

let's find the new linear density

μ = m / l

iinitial density

μ₀ = m₀ / l

final density

μ = 2m₀ / lo

μ = 2 μ₀

we substitute in the equation for the velocity

initial v₀ = $\sqrt{ \frac{T_o}{ \mu_o} }$

with the new dough

v = $\sqrt{ \frac{T_o}{ 2 \mu_o} }$

v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }

v = 1 /√2 v₀

v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

μ = μ₀

in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

μ = 3m₀ / l₀

μ = 3 m₀/l₀

μ = 3 μ₀

v = $\sqrt{ \frac{T_o}{ 3 \mu_o} }$

v = 1 /√3 v₀

v = 0.577 v₀

d) l = 2l₀

μ = m₀ / 2l₀

μ = μ₀/ 2

v = $\sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }$

v = √2 v₀

v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

μ = 10 m₀/2l₀

μ = 5 μ₀

we look for speed

v = $\sqrt{ \frac{T_o}{5 \mu_o} }$

v = 1 /√5 v₀

v = 0.447 v₀

5 0

3 years ago

Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to

lbvjy [14]

Answer:

Explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁ - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

v = 1.1 m /s

Tan θ = .76 / .8

θ = 44°

6 0

3 years ago

Because the Triple Beam Balance is zeroed before being put into use, it measures

pickupchik [31]

Answer:

B

Explanation:

8 0

3 years ago