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3 years ago

A(n) 59 kg astronaut becomes separated from

Physics

1 answer:

pantera1 [17]3 years ago

5 0

Answer:

5. 7.452 min

Explanation:

Momentum is conserved, so:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

0 = (0.695 kg) (-12 m/s) + (59 kg) v

v = 0.1414 m/s

She is 63.2 m away, so the time it takes to reach the shuttle is:

t = 63.2 m / 0.1414 m/s

t = 447.1 s

t = 7.452 min

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He does whatever a spider can. Spider-Man, who has a mass of 76 [kg], is clinging onto an inclined wall forming an inclination a

Aliun [14]

Answer:570.54 N

Explanation:

Given

mass of man=76 kg

$\theta =50^{\circ}$

As man is standing over inclined building therefore

its weight has two components i.e. sin and cos component

Force perpendicular to inclined wall

$F=mgcos\theta =76\times 9.8\times \sin 50$

F=570.54 N

4 0

3 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$

The time independent constant is 16.0787 s

$E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}$

At t = 6

$\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}$

From the first equation

$\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451$

$E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa$

$E_r(6)=4.35614\ MPa$

6 0

3 years ago

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick

Travka [436]

Answer:

c. vf is greator than v2, but less than v1

Explanation:

The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.

In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.

Collisions in which the kinetic energy is conserved are called elastic collision.

Collisions in which the kinetic energy is not conserved are called inelastic collisions. If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.

<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>

Therefore,

a. vf = v2 is wrong because vf is greater than v2

b. vf is less than v2 is wrong because vf is greater than v2

c. vf is greater than v2, but less than v1 is correct.

d. vf = v1 is wrong because vf is less than v1

4 0

3 years ago

Where is UTC the local time?

Dmitrij [34]

UTC is generally adopted as the local time at Prime Meridian (PM).

UTC is an abbreviation for Coordinated Universal Time and it can be defined as a local time standard by which all clocks, time and time zones are regulated around the world.

This ultimately implies that, Coordinated Universal Time (UTC) is generally adopted as the local time at Prime Meridian (PM) because it nominally reflects mean solar time.

Read more on UTC here: brainly.com/question/14333785

#SPJ12

5 0

2 years ago

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m

son4ous [18]

Answer:

A. 30.7cm

B. $1.7*10^{-10}C$

C. The electric field is directed away from the point of charge

Explanation:

$because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm$

Considering Gauss's law

$EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C$

C. The electric field directed away from the point of charge when the charge is positive.

3 0

3 years ago