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3 years ago

Can someone answer these questions for me all 10 I’m confused

Physics

1 answer:

satela [25.4K]3 years ago

7 0

Answer:

Is this the whole paper? I'll need to know so I can answer.

Explanation:

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Calculate the centripetal force acting on a 925 kg car as it rounds an unbanked curve with a radius of 75 m at a speed of 22 m/s

igomit [66]

Since f=m(v^2/r),or fnet is equal to ma.

force = unknown
velocity=22m/s
radius=75m

f=m(v^2/r)
f=925(22^2/75)
f=5969.333N

3 0

3 years ago

Which of the following means that an image is located behind a lens? A. +di B. -di C. -do D. +do

STatiana [176]

For a lens, the following sign convention is generally used:

- f (the focal length) is positive for a converging lens and negative for a diverging lens
- $d_o$ (the distance of the object from the lens) is positive if the object is in front of the lens and negative if it is behind the lens
- $d_i$ (distance of the image from the lens) is positive if the image is behind the lens (real image) and negative if the image is in front of the lens (virtual image)

Therefore, the correct option is
<span>A. +di
</span>which mens that the image is real and located behind the lens.

4 0

3 years ago

A planet orbits a star, in a year of length 2.27 x 107 s, in a nearly circular orbit of radius 3.99 x 1011 m. With respect to th

luda_lava [24]

Explanation:

It is given that,

Radius of circular orbit, $r=3.99\times 10^{11}\ m$

Time taken, $t=2.27\times 10^{7}\ m$

(a) Angular speed of the planet is given by :

$\omega=\dfrac{2\pi}{t}$

$\omega=\dfrac{2\pi}{2.27\times 10^{7}\ m}$

$\omega=2.76\times 10^{-7}\ rad/s$

(b) The tangential speed of the planet is given by :

$v=r\times \omega$

$v=3.99\times 10^{11}\times 2.76\times 10^{-7}$

v = 110124 m/s

(c) The centripetal acceleration of the planet,

$a=\dfrac{v^2}{r}$

$a=\dfrac{(110124)^2}{3.99\times 10^{11}}$

$a=0.0303\ m/s^2$

Hence, this is the required solution.

3 0

3 years ago

A block of mass 0.252 kg is placed on top of a light, vertical spring of force constant 4 825 N/m and pushed downward so that th

svp [43]

Answer:

10.77m

Explanation:

The elastic potential energy of the spring when compressed with the mass is converted to the kinetic energy of the mass when released. This ie expressed in the following equation;

$\frac{1}{2}ke^2=\frac{1}{2}mu^2..............(1)$

where k is the force constant of the spring, e is the compressed length, m is the mass of the block and u is the velocity with which the block leaves the spring after being released.

If we make u the subject of formula from equation (1) we obtain the following;

$u=\sqrt{\frac{ke^2}{m}}................(2)$