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aniked [119]
3 years ago
10

A 1200 kg car is moving at 5.0 m/s east. it strikes an 1800 kg car at rest. the cars have an elastic collision and move in the e

ast or west direction. the velocity of the 1200 kg car after the collision is:
Physics
1 answer:
spayn [35]3 years ago
5 0
Velocity = (1200*5)/(1200+1800)=2
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What is the length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.2
Deffense [45]

Answer:

Option B is the correct answer.

Explanation:

Thermal expansion

            \Delta L=L\alpha \Delta T

L = 1.2 meter

ΔT = 65 - 15 = 50°C

Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C

We have change in length

          \Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m

New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m

Option B is the correct answer.

3 0
3 years ago
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A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the batte
Assoli18 [71]

Answer:

<em>There will be an increase in potential difference.</em>

Explanation:

As we know that the potential difference depends upon the capacitance.

ΔV = Q/C

When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.

In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.

4 0
3 years ago
John goes grocery shopping with his mother. His job is to push the cart. The cart is
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Answer:

Beacause he has more grocceries and food heavy

Explanation:

7 0
3 years ago
The top of the cylinder head is sealed by the
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The common name is valve cover, so the correct answer is A.
3 0
3 years ago
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An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
2 years ago
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