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finlep [7]
3 years ago
7

Suppose that on a routine trip around the galaxy, your dog puts his head out of the spacecraft window (as dogs often do). The so

lar wind has a temperature of approximately 104 K. Would your dog’s head burn up immediately? Explain briefly in the context of temperature versus heat (assume your dog holds his breath during the exercise and can withstand low pressure).
Physics
1 answer:
neonofarm [45]3 years ago
7 0

Answer: Yes, the dog's head will burn

Explanation:Solar winds are strongly destructive magnetically charged high energy winds. Satellite communication on earth and the occasional geomagnetic storm near the poles are the major disruptive effects caused by solar winds on earth.

Since solar winds are streams of charged particles released from the upperparts of the Sun, called the "corona". This plasma mainly consists of electrons, protons and alpha particles with kinetic energy between 0.5 and 10 keV. Instilled within the solar-wind plasma is the interplanetary magnetic field.

If the dog holds it's breath the heat wave would burn it's head but not outrightly because the breath has ceased for some seconds and it can't absorb the much heat of the solar wind into it's nostrils. The effect of holding its breath would still be disastrous but not as the first case.

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A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
4 years ago
he acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. The
Ahat [919]

1/3 the weight than it is on earth

5 0
4 years ago
Twenty is the _________________ of potassium.
Orlov [11]

Answer:

I think it's D) Guess why? I searched it up!

6 0
3 years ago
Read 2 more answers
Particle A of charge 3.06 10-4 C is at the origin, particle B of charge -5.70 10-4 C is at (4.00 m, 0), and particle C of charge
Alex

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

F_ac =  8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc =  8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2  

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

5 0
4 years ago
A block of mass M1 = 288.0 kg sits on an inclined plane and is connected to a bucket with a massless string over a massless and
AlekseyPX

Answer:

M2 = 278.06 kg

Explanation:

We calculate the weight of M1

W=m*g

Where

m: mass (kg)

g: acceleration due to gravity (m/s²)

W₁=288* 9.8= 2822.4 N

Look at the attached graphic

We calculate the x-y components of the weight :

W₁x= 2822.4*sin41° N =1851.66 N

W₁y= 2822.4 *cos41° N = 2130.09 N

We apply Newton's first law for the balance in M1:

Σ Fy=0

Fn-W₁y=0  ,   Fn: normal force

Fn=W₁y=2130.09N

Friction Force = Ff=μs *Fn = 0.41*2130.09 =873.34 N

Σ Fx=0

T- W₁x- Ff=0

T= 1851.66 + 873.34

T= 1851.66 + 873.34

T=2725 N

We apply Newton's first law for the balance in M2:

Σ Fy=0

T- W₂ =0

W₂ = T = 2725 N

W₂ = M2*g

M2 = W₂/g

M2 = 2725/9.8

M2 = 278.06 kg

6 0
3 years ago
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