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3 years ago

4. A roller coaster starts at the top of a hill, 42 m above ground, with a velocity of 5.0 m/s.

Physics

1 answer:

Lady bird [3.3K]3 years ago

3 0

Answer:

A roller coaster starts from rest at a point 45 m above the bottom of a dip (See Fig. 6-2). Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 30 m above the bottom of the dip?

a. 17 m/s b. 24 m/s c. 14 m/s d. 30 m/s

You have no mass given, so you work it cancelling the mass:

i'll take g = 9.8 m/s^2

(g x h1) - (g x h2) = 1/2v^2 ( no mass either for KE)

(9.8 x 45) - (9.8 x 30) = 147

147 x 2 = v^2 = 294, sq-rt = v = 17.146 m/s answer (17m/s)

A roller coaster starts with a speed of 5.0 m/s at a point 45 m above the bottom of a dip (See Fig. 6-2). Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 30 m above the bottom of the dip? a. 16 m/s b. 12 m/s c. 14 m/s d. 18 m/s

u^2 = 2gh (5^2 = 2gh)

25/(2g) = h = 1.2755 m

45 + 1.2755 = 46.2755m

Do the same equations as question one, answer = 17.86 m/s (answer 18 m/s)

A roller coaster starts at a point 30 m above the bottom of a dip with a speed of 25 m/s (See Fig. 6-2). Neglect friction, what will be the speed of the roller coaster at the top of the next slope, which is 45 m above the bottom of the dip? a. 14 m/s b. 16 m/s c. 20 m/s d. 18 m/s

Do the same! 25^2 = 2gh, 625/(2g) = h = 31.89 m

31.89 + 30 = 61.89 m

(9.8 x 61.89) - (9.8 x 45) = 1/2v^2 = 165.52

165.52 x 2 = 331.04, sq-rt = 18.19 m/s (answer 18 m/s)

A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart?

a. 73 m/s b. 6.0 m/s c. 36 m/s d. 8.5 m/s

Ignoring friction:

F = ma

200/55 = a = 3.636 m/s^2

v^2 = u^2 + 2as, u^2 = zero

v^2 = 2as

2as = 2 x 3.636 x 10 = 72.72, sq-rt = 8.52 m/s (answer 8.5 m/s)

Explanation:

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A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .

As a result,

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V$ .

9 0

3 years ago

A current of 0. 82 a flows through a light bulb. how much charge passes through the light bulb during 94 s?

gladu [14]

A current of 0. 82A flows through a light bulb. The charge passed through the light bulb during 94 s is 77.08C

The amount of charge flown for a given period of time determines the current passed through a bulb or electrical body.

The relation between the charge, current and time is given as:

Q = I × t

where, Q is the charge flown through bulb

I is the current passed through bulb

t is the time for which charge passes through bulb

Given,

I = 0.82A

t = 94s

Q = ?

Substituting the values in the above formula:

Q = I × t

Q = 0.82 × 94

Q = 77.08C

Hence, The charge passed through the light bulb during 94 s is 77.08C

Learn more about Current here, brainly.com/question/2264542

#SPJ4

4 0

2 years ago

For problems 2-3, a Ferrari accelerates from 0-60.0 miles per hour in 2.50 seconds.

Montano1993 [528]

Answer:

Explanation:

The first part of this question is simply asking us to convert the speed from miles per hour to meters per second:

$60.0\frac{mi}{hr}*\frac{1hr}{3600sec}*\frac{1609.34m}{1mi}=26.8\frac{m}{s}$ choice C.

The next part wants us to use the equation for acceleration and find the acceleration:

$a=\frac{v-v_0}{t}$ where v is final velocity, v0 is initial velocity, and t is time in seconds (which was one of the reasons we had to convert the initial velocity from 60.0 mph to m/s):

$a=\frac{26.8-0}{2.5}$ and

a = 10.7 m/s/s, choice B.

6 0

3 years ago

Discuss how lawmakers could use scientific knowledge to help them make decisions that would benefit society.

Anika [276]

Scientific knowledge is very firm in its facts and pieces of evidence that create a more precise definition. If lawmakers are able to use the scientific method in making a decision, it will make a big change and will be beneficial to the society. The set of rules will be more arranged, specific and restrictive that might change justice, tax, and law.

7 0

3 years ago

Determine the stretch in OA spring required to hold the 21kg- crate in the equilibrium position shown.

Fynjy0 [20]

The stretch of the OA spring is calculated by the formula of F=-k*(extension) and thats the formula for elasticity. Based on my calculation i came up with an answer of the unstretched length is 2m abd the stiffnes is 310n/m

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4 years ago