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2 years ago

A roller coaster is stationary at the top of a ramp.

Physics

1 answer:

lora16 [44]2 years ago

3 0

I think that the answer is B.

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The motorcycle travels with a constant speed v0 along the path that, for a short distance, takes the form of a sine curve. Deter

Alexandra [31]

Let at any instant of time the speed is vo and the angle made by the bike with the horizontal is given

now we have

component of speed in x direction given as

$v_x = v_0cos\theta$

component of speed in y direction will be

$v_y = v_0sin\theta$

now from above two equations we can say that here

$\theta$ = angle with the horizontal at any instant

and since here it is a sine curve so we know that

$y = sin(x)$

so we have slope of graph

$tan\theta = \frac{dy}{dx} = cos(x)$

6 0

2 years ago

When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin

saveliy_v [14]

True
If it helps you plz brainlest me

7 0

2 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

2 years ago

Speed is a scalar quantity that describes only the

Anna [14]

Answer:

magnitude of the velocity

Explanation:

4 0

2 years ago

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

$W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m$

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

$W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J$

So, the new work is more than 130 J.

6 0

2 years ago