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Vaselesa [24]
4 years ago
9

The first hill of a roller coaster is 42 meters high. The top of the second hill of the one

Physics
2 answers:
AleksAgata [21]4 years ago
8 0

Answer:

Explanation:

The first hill of a roller coaster is 42 meters high. The top of the second hill of the roller coaster is 33 meters high. The mass of the roller coaster is 34,500 kg

Marta_Voda [28]4 years ago
7 0

Answer:

Part a)

Speed of the roller coaster is

v = 13.3 m/s

Part b)

Since it is moving with non zero speed at some height above the ground

So we will have

Kinetic energy + Potential energy Both

Explanation:

As we know that there is no friction on the path

So here we can use mechanical energy conservation law

so we will have

Part a)

U_i  + K_i = U_f + K_f

mgh + 0 = \frac{1}{2}mv^2 + mgH

34500(9.81)(42) = \frac{1}{2}(34500)v^2 + 34500(9.81)(33)

v = 13.3 m/s

Part b)

Since it is moving with non zero speed at some height above the ground

So we will have

Kinetic energy + Potential energy Both

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A radioactive decay is illustrated. Which radioactive emission is a product of this decay?
hram777 [196]

The radioactive decay is shown in the image attached to this answer

Answer:

Alpha

Explanation:

An alpha decay is said to have occurred when the mass number of the daughter nucleus decreases by four units and the atomic number of the daughter nucleus decreases by two units.

If we look at the equation from left to right, we will notice that the mass number of the daughter nucleus decreased by four units and the atomic of the daughter nucleus number by two units compared to that of the parent nucleus leading to the emission of an alpha particle.

5 0
3 years ago
A ball is being thrown straight upward and rises to a maximum height of 12.0m above its launch point. At what height above it’s
AlexFokin [52]

Answer:

9.0 m

Explanation:

Let the initial velocity be 'u'.

Given:

Final velocity is half of initial velocity.

Maximum height reached by ball (H) = 12.0 m

Acceleration of the ball is due to gravity (g) = -9.8 m/s²(Downward)

Now, first, we will find initial velocity of the ball using equation of motion given as:

v^2=u^2+2a(\Delta y)

For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So, v=0\ m/s

Also, \Delta y=H=12\ m

Now, plug in all the values and solve for 'u'.

0^2=u^2+2(-9.8)(12)\\\\u^2=235.2\\\\u=\sqrt {235.2} =15.34\ m/s

Now, consider the motion of the ball till the velocity reaches half of initial velocity.

So, final velocity (v) = \frac{u}{2}=\frac{15.34}{2}=7.67\ m/s

Now, again using the same equation and finding the new height now. Let the new height be 'h'.

So, equation of motion is given as:

v^2=u^2+2ah\\7.67^2=15.34^2+2\times -9.8\times h\\58.83=235.2-19.6h\\\\19.6h=235.2-58.83\\\\19.6h=176.37\\\\h=\frac{176.37}{19.6}\approx9.0\ m

Therefore, the height reached by the ball when velocity is decreased to one-half of the initial velocity is 9.0 m.

8 0
3 years ago
The length of the assembly decreases by 0.006 in. when an axial a- force is applied by means of rigid end plates. Determine
Triss [41]

Answer:

(a) The magnitude of the applied force is (0.0001524k) Newton

(b) Corresponding stress in the steel core = (0.0001524k/area) Newton per meter square

Explanation:

(a) From Hookes law of elasticity,

Force applied = force constant (k) × compression

compression = 0.006 in = 0.006 × 0.0254 = 0.0001524 meter

Force applied = k × 0.0001524 = (0.0001524k) Newton

(b) Stress = Force applied (Newton)/area of steel core (meter square) = (0.0001524k/area) Newton per meter square

6 0
3 years ago
If you drive at 50 km/hr for 8 hours, how far will you go?
bonufazy [111]
50km/hr* 8hr = 400 km
7 0
3 years ago
Why did the model spacecraft go so much faster than expected on Wednesday?
Andrew [12]

Answer:

The Wednesday test launch stored more potential energy, and launched the spacecraft at a faster speed because the stronger magnetic field closer to the magnet resulted in a greater increase in potential energy.

Explanation:

6 0
3 years ago
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