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Charra [1.4K]
3 years ago
12

A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i

s the change in momentum of the car?
Physics
1 answer:
antiseptic1488 [7]3 years ago
4 0

Answer:

The change in momentum = -20000 kg m/s.

Explanation:

Mass m = 1000 kg

speed v₁ = 20 m/s

speed v₂ = 0 m/s

We know that,

The change in momentum

ΔP = m (Δv)

ΔP = m (v₂ - v₁)

     = 1000 (0 - 20)

     = 1000 (-20)

     = -20000 kg m/s

Thus, the change in momentum = -20000 kg m/s.

Note: negative sign indicates that the velocity is reducing when it hits the barrier.

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Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of
sp2606 [1]

Answer:

Explanation:

Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass  2.3 x 10¹⁹ Kg

Final moment of inertia I₂ = 2/5 M R² + 2/3  x 2.3 x 10¹⁹ x R²

For change in period of rotation we shall apply conservation of angular momentum law

I₁ ω₁  = I₂ ω₂  ,  ω₁ and   ω₂ are angular velocities initially and finally .

I₁ / I₂     =  ω₂ / ω₁

I₁ / I₂     =  T₁ / T₂  , T₁ , T₂ are time period initially and finally .

T₂ / T₁ = I₂ / I₁

(2/5 M R² + 2/3  x 2.3 x 10¹⁹ x R²) / 2/5 MR²

1 + 5 / 3  x 2.3 x 10¹⁹ / M

= 1 + 5 / 3  x 2.3 x 10¹⁹ / 5.97 x 10²⁴

= 1 + .0000064

T₂ = 24 (1 + .0000064)

= 24 hours + .55 s

change in length of the day = .55 s .

3 0
3 years ago
Compare the two numbers using <, >, =
Effectus [21]
Vas happenin!!

1. =
2. >
3. <
4. <
5. <
6. >

Hope this helps *smiles*

3 0
3 years ago
An astronaut on the moon places a package on a scale and finds its weight to be 10. N. () What would the weight of the package
jeka94

Answer:

(a) 61.25 N

(b) 6.25 kg

(c) 6.25 Kg

Explanation:

Weight on moon = 10 N

Acceleration due to gravity on moon = 1.6 m/s^2

Acceleration due to gravity on earth = 9.8 m/s^2

Let m be the mass of the package.

(a) Weight on earth = mass x acceleration due to gravity on earth

Weight on earth = 6.25 x 9.8 = 61.25 N

(b) Weight on moon = mass x acceleration due to gravity on moon

10 = m x 1.6

m = 6.25 kg

(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.

8 0
3 years ago
A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
Montano1993 [528]
<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

g=-9.8\frac{m}{s^{2}} is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}   (9)

-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
What do you mean by booting ?whst are typez of booting​
Evgen [1.6K]

Answer:

plz mark brainlist

8 0
3 years ago
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