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AveGali [126]
3 years ago
10

PLEASE HELP ME WHATS THE ANSWER ???

Physics
1 answer:
otez555 [7]3 years ago
8 0

Answer:

A

Explanation:

VIRTUAL, UPRIGTH AND REDUCED

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A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
F_a=m\frac{v^2}{r}-mg=179 $N
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}

6 0
3 years ago
A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate
schepotkina [342]

Answer:

Explanation:

impedance z=(XL^2+R^2)^1/2

power across te resistor ==i^2r

286/300

I=.976

3 0
3 years ago
A kite 100 ft above the ground moves horizontally at a speed of 12 ft/s. at what rate is the angle (in radians) between the stri
Viefleur [7K]
<span>Answer: So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z. So we have x2 + 1002 = z2. Now take its derivative in terms of time to get 2x(dx/dt) = 2z(dz/dt) So at your specific moment z = 200, x = 100âš3 and dx/dt = +8 substituting, that makes dz/dt = 800âš3 / 200 or 4âš3. Part 2 sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get cos a (da./dt) = -100/ z2 (dz/dt) So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or Ď€/6 radians. Substitute to get cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3) âš3 / 2 (da/dt) = -âš3 / 100 da/dt = -1/50 radians</span>
5 0
3 years ago
________ is an arrhythmia in which there is a very fast but regular rhythm (250 beats per minute) of the atria or ventricles.
erastova [34]

Answer:

Flutter

Explanation:

Flutter is a type of arrhythmia that causes very fast and regular ryth of the atria of about 250 beats per minute.

Arrhythmia can be defined as any sort of irregularity heart rate or rhythm is also called as dysrhythmia.

Arrhythmias can be categorized as heart block, bradycardia, tachycardia, fibrillation, flutter, sick sinus syndrome, and is diagnosed by Electrocardiography.

In Flutter, the heart chambers do get sufficient time to get filled with blood completely prior to next contraction.

4 0
3 years ago
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]

where

\phi = Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

\lambda = wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]

Magnetic field Equation

B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]

Where B₀= E₀/c

B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T

B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]

6 0
3 years ago
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