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2 years ago

Which is NOT an effect of friction?

Physics

1 answer:

Bezzdna [24]2 years ago

6 0

Answer:

The answer is add mass.

Explanation:

Friction does NOT add mass to an object. It does generate heat, slow objects, and wear down surfaces.

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The value of g is greater at the poles than at the equator why

babunello [35]

Answer:

because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

6 0

2 years ago

Give three factors which are responsible for the vanishing forest

slava [35]

Answer:

1. Huge wildfires

2. Deforestation

3. Reduced amount of aforestation, etc

5 0

2 years ago

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5

Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

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

where $F_T = 44N$ is the tension on the rope and $\mu = m/L = 0.5 / 2.6 = 0.1923 kg/m$ is the density of the rope per unit length

$v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s$

6 0

2 years ago

When would the carrying capacity of an area be most likely to change? Choose the correct answer.

pickupchik [31]

Answer:

when resources remain the same

Explanation:

The capacity of an area will change if the available resources remain the same, especially if the population is changing.

8 0

2 years ago

A toy car having mass m = 1.50 kg collides inelastically with a toy train of mass M = 3.60 kg. Before the collision, the toy tra

baherus [9]

Answer:

$v_{f}$ = 3,126 m / s

Explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

p₀ = m v₁₀ + M v₂₀

After the inelastic shock

$p_{f}$ = m $v_{1f}$ + M $v_{2f}$

p₀ = $p_{f}$

m v₀ + M v₂₀ = m $v_{1f}$ + M $v_{2f}$

We cleared the end of the train

M $v_{2f}$ = m (v₁₀ - v1f) + M v₂₀

Let's calculate

3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

$v_{2f}$ = (-3.9 + 8.82) /3.60

$v_{2f}$ = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

m v₁₀ + M v₂₀ = (m + M) $v_{f}$

$v_{f}$ = (m v₁₀ + M v₂₀) / (m + M)

$v_{f}$ = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

$v_{f}$ = 3,126 m / s

4 0

2 years ago