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2 years ago

You should be particularly careful a restraining a struggling rabbit because it can

1 answer:

6 0

You have no options here so I'll just answer. It can cause a rise in heart rate and greatly increases the risk of overheating and even death. If you grab the rabbit too hard, you risk breaking/fracturing a bone or causing other kinds of damage, whether externally or internally, to the rabbit.

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What are the very small particles that make up matter

mrs_skeptik [129]

Atoms
Hope this helps!!

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2 years ago

Read 2 more answers

An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol

iogann1982 [59]

Answer:

(a) B = 2.85 × $10^{-6}$ Tesla

(b) I = I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = $\frac{mv}{qB}$

⇒ B = $\frac{mv}{qr}$

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = $\frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }$

= $\frac{9.11*10^{-27} }{3.2*10^{-21} }$

B = 2.85 × $10^{-6}$ Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒ I = B ÷ μN/L

where B is the magnetic field, μ is the permeability of free space = 4.0 × $10^{-7}$ Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

= $\frac{2.85*10^{-6} }{4*10^{-7} *25}$

I = 0.285 A

5 0

2 years ago

What causes a snowflake to melt on your tongue?

Elodia [21]

Answer:

thermal energy

Explanation:

heat transfers into it causing it to physically change

7 0

2 years ago

Read 2 more answers

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

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2 years ago

Physics Circuit, need a walkthrough

QveST [7]

I don't see a picture

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2 years ago