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3 years ago

Describe a real-world example of the law of conservation of momentum.

1 answer:

4 0

Imagine a car crash. A car coming at a high speed has a head on collision with a car at rest. When the car makes impact, it will move the other car with it at a slower speed then it was travelling at. In this case, the velocity decreased since the car slowed down, but the mass increased since there are now two cars moving. Momentum was conserved because the change in mass accounts for the loss of velocity.

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Words a child uses are called the spoken or ___ vocabulary

Vaselesa [24]

The answer is: Expressive vocabulary

Expressive vocabulary refers to the combination of all the words that a person has acquire throughout his/her life and can be used in various type of situations.

This would include all words in the child vocabulary, starting from the child's written language, spoken language or even the child's manually signed words.

4 0

3 years ago

A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass

m_a_m_a [10]

Answer:

F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

$F=G\dfrac{m_1m_2}{R^2}$

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

$F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N$

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

5 0

3 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago

A 0.0414 kg ingot of metal is heated to 243◦C

tino4ka555 [31]

Answer:

448 J/kg/°C

Explanation:

m₁ C₁ (T₁ − T) + m₂ C₂ (T₂ − T) = 0

(0.0414 kg) C (243°C − 20.4°C) + (0.411 kg) (4186 J/kg/°C) (18°C − 20.4°C) = 0

(9.22 kg°C) C − 4129 J = 0

C = 448 J/kg/°C

3 0

3 years ago

The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.0

Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If, V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t

a(0) = 1.5m/s^²

1.5m/s^² = A + 2B × 0

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²

1.5× 1 +B× 1 = 2m/s

B = 2-1.5

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

7 0

4 years ago