All categories

3 years ago

As a roller coaster car crosses the top of a 50-m-diameter loop-the-loop, its apparent weight is the same as its true weight.

Physics

1 answer:

scZoUnD [109]3 years ago

3 0

Answer:22.36 m/s

Explanation:

Given

the diameter of loop d=50 m

the radius of loop r=25 m

At the top position, we can write,

weight and Normal reaction combination will provide the centripetal force i.e.

$R+W=\frac{mv^2}{r}$

$R=W\quad \quad [\text{apparent weight =Actual weight}]$

$2W=2mg=\frac{mv^2}{r}$

$v=\sqrt{2gr}$

$v=\sqrt{2\times 10\times 25}$

$v=22.36\ m/s$

You might be interested in

figure 2 shows a charged ball of mass m = 1.0 g and charage q = -24*10^-8 c suspended by massless string in the presence of a un

Vlad [161]

Answer:

E = 307667 N/C

Explanation:

Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.

This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:

0.0098 N = T * cos(37)

then T = 0.0098/cos(37) N = 0.01227 N

Knowing the tension's magnitude, we can find its horizontal component:

T * sin(37) = 0.007384 N

and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:

0.007384 N = E * 2.4 * 10^(-8) C

Then E = 0.007384/2.4 * 10^(-8) N/C

E = 307667 N/C

8 0

3 years ago

During the annual shuffleboard competition, Renee gives her puck an initial speed of 8.12 m/s. Once leaving her stick, the puck

shutvik [7]

<h2>Answer:2.65 seconds</h2>

Explanation:

Let $a$ be the acceleration.

Let $u$ be the initial velocity.

Let $v$ be the final velocity.

Let $t$ be the time taken.

As we know from the equations of motion,

$v=u+at$

Given, $v=0\\u=8.12ms^{-1}\\a=-3.06ms^{-2}$

$0=8.12-3.06t$

$t=2.65sec$

6 0

3 years ago

What happens to the acceleration of a ball as it is dropped off a clift?

denpristay [2]

Answer:

become positive

Explanation:

6 0

3 years ago

Two coils close to each other have a mutual inductance of 32 mH. If the current in one coil decays according to I=I0e−αt, where

fiasKO [112]

The emf induced in the second coil is given by:

V = -M(di/dt)

V = emf, M = mutual indutance, di/dt = change of current in the first coil over time

The current in the first coil is given by:

i = i₀ $e^{-at}$

i₀ = 5.0A, a = 2.0×10³s⁻¹

i = 5.0e^(-2.0×10³t)

Calculate di/dt by differentiating i with respect to t.

di/dt = -1.0×10⁴e^(-2.0×10³t)

Calculate a general formula for V. Givens:

M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)

Plug in and solve for V:

V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))

V = 320e^(-2.0×10³t)

We want to find the induced emf right after the current starts to decay. Plug in t = 0s:

V = 320e^(-2.0×10³(0))

V = 320e^0

V = 320 volts

We want to find the induced emf at t = 1.0×10⁻³s:

V = 320e^(-2.0×10³(1.0×10⁻³))

V = 43 volts

3 0

3 years ago

Carlos wanted to know how positive and

Iteru [2.4K]

Answer:

it Give only one of them a positive or negative charge

5 0

2 years ago

Read 2 more answers