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2 years ago

Pls Help Me with this problem

Physics

2 answers:

Naddika [18.5K]2 years ago

8 0

Answer:

explanation

Explanation:

1 = C

2 = A

3 = D

4 = E

5 = B

allsm [11]2 years ago

5 0

1) C, 2) B, 3) D, 4)A, 5)E

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Krishne wants to measure the mass and volume of a thimble. Which tools should she use?

andrew11 [14]

To measure the mass, you would use a balance. To measure the volume, you can use a variety of ounces, cups, pints, quarts, and gallons. Good luck!

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3 years ago

A substance that does not react with metal, has a bitter taste, and turns litmus paper blue is a(n) ___.

Pavlova-9 [17]

The answer to the question is b because acid is a substance that tastes sour, reacts with metal and turns litmus paper blue

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3 years ago

A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a

trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden $d_1=1.1cm$

Speed of water in the hose is $v_1=0.95m/sec$

Number of holes n = 22

Diameter of each holes $d_2=15cm$

According to continuity equation $A_1v_1=A_2v_2$

$d_1^2\times v_1=22\times d_2^2v_2$

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6 0

3 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ (we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) what fraction of its initial energy is los

Scilla [17]

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100

and since mass and gravity are constant so it leaves us with just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the energy or 0.25 in fraction

b) Here use the equation vf^2=vi^2+2gd

where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable
we know that at maximum height or peak, the velocity is 0 so vf is 0

therefore,

vi =sqrt(-2gd) 
vi =sqrt(-2x-9.81x1.5) 
vi =5.4 m/s

c) The energy was converted to heat due to friction with the air and the ground.

6 0

3 years ago