Ask question

Ask question

All categories

2 years ago

15

A rabbit trying to escape a fox runs north for 8.0 m, darts northwest for 1.0 m, then drops 1.0 m down a hole into its burrow. W

hat is the magnitude of the net displacement of the rabbit?

1 answer:

Andrews [41]2 years ago

7 0

Answer: The net displacement is 6.40meters

Explanation:

You might be interested in

The CERN particle accelerator is circular with a circumference of 7.0 km.

Contact [7]

Answer:

$a_c=2.0196\times 10^{13}\ m/s^2$

$F=3.37273\times 10^{-14}\ N$

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

v = Speed of proton = 0.5c = $0.5\times 3\times 10^8=1.5\times 10^8\ m/s$

Circumference of the colider is 7 km

$P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m$

$a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2$

Centripetal acceleration is $2.0196\times 10^{13}\ m/s^2$

$F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N$

Force on protons is $3.37273\times 10^{-14}\ N$

8 0

3 years ago

How are metals identified i the periodic table??

zaharov [31]

Answer:

okay here is a thing I learned when I was younger in my middle school:

Explanation:

my teacher would tell me that metals are considered a weak metals are on the left side and the good metals are located on the right side because the only way I remembered was the right means it is really strong and the left is weak and not that supportive. but I think that's how I still think it is or other people may have their own opinions. but hope this helped out with your question!

8 0

3 years ago

Read 2 more answers

A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate

babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

$\phi = NBA$

here we have

$A = \pi R_s^2$

$B = \mu_o N i/L$

now we have

$\phi = N(\mu_o N i/L)(\pi R_s^2)$

now the induced EMF is rate of change in magnetic flux

$EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

now for induced electric field in the coil is linked with the EMF as

$\int E. dL = EMF$

$E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

$E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}$

$E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}$

$E = 58.7 V/m$

3 0

2 years ago

Question 1 of 20

ad-work [718]

Answer

D. move a small magnet back and forth within a section of the coiled wire.

Explanation:

i put that for the test and i got it right

8 0

2 years ago

Read 2 more answers

Ozone, O3 methane, CH4 oxygen gas, O2 Are classified as

erma4kov [3.2K]

I would think compounds.

5 0

2 years ago

Read 2 more answers