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3 years ago

Hey besties, im having a mental breakdown on this question. pls help :D

1 answer:

8 0

With "toward the wall" designated as the positive direction, the average acceleration felt the ball in the 0.0158 s of contact with the wall is

$a_{\rm ave} = \dfrac{-26.412\frac{\rm m}{\rm s} - 28.4\frac{\rm m}{\rm s}}{0.0158\,\mathrm s} ≈ \boxed{-3470 \dfrac{\rm m}{\mathrm s^2}}$

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What type of weather would be in the middle of the picture, where the barometric pressure is 1000 millibars?

UkoKoshka [18]

It would be D. Hope that helped :)

4 0

3 years ago

Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f

RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

$p_A = p_B$

Pressure is given by the ratio between force F and area A, so we can write

$\frac{F_A}{A_A}=\frac{F_B}{A_B}$

The force exerted on each piston is just equal to the weight of the corresponding mass: $F=W=mg$ , where m is the mass and g is the gravitational acceleration. So the equation becomes

$\frac{m_A g}{A_A}=\frac{m_B g}{A_B}$

Now we can rewrite the mass as the product of volume, V, times density, d:

$\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}$

We also know that $A_B = 2.0 m^2\\A_A = 1.0 m^2$

So we can further re-arrange the equation (and simplify g as well):

$\frac{V_A d_A}{1}=\frac{V_B d_B}{2}$

$\frac{d_A}{d_B}=\frac{V_B}{2V_A}$

We are also told that block B has bigger volume than block A: $V_B > V_A$ . However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0

3 years ago

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm

iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

$m = m_{o}* m_{e}$

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

$m_{o} = \frac{v_{o}}{-u _{0}}$

and me as

$m_{e} = 1+\frac{D}{f_{e}}$

d is distant of distinct vision = 25.0 cm for normal eye

fe = focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}$

$\frac{1}{v_{o}} = 2.1 cm$

$m_{o} = \frac{2.1}{0.150} = 14$

$m_{e} = 1+\frac{25}{1}$

$m_{e} =26$

$m = m_{o}* m_{e}$

m = 14*26 = 364

4 0

3 years ago

uphill at a rate of 2.5 mi/h from the base of a 6-mi trail. At the same time, Edwin walks downhill at a rate of 3.5 mi/h from th

uranmaximum [27]

<h2>After 1 hour they meet.</h2>

Explanation:

Distance between them = 6 miles

Speed of uphill person = 2.5 miles per hour

Speed of downhill person = 3.5 miles per hour

Relative velocity = 2.5 - ( -3.5 ) = 6 miles per hour

We know

Displacement = Velocity x Time

6 = 6 x Time taken

Time taken = 1 hour

After 1 hour they meet.

4 0

4 years ago

Is space expanding within clusters of galaxies? why or why not?

GarryVolchara [31]

No, since their gravity is powerful enough to keep them together even while the universe expands as a whole. Space is not expanding within clusters of galaxies.

<h3>What is a galaxy?</h3>

A galaxy is a massive clump of gas, dust, and billions of stars and their solar systems bound together by gravity.

No, since their gravity is powerful enough to keep them together even while the universe expands as an entire.

Hence,space is not expanding within clusters of galaxies.

To learn more about the galaxy, refer to the link;

brainly.com/question/2905713

#SPJ1

6 0

2 years ago