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iren2701 [21]
3 years ago
5

The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.

Physics
1 answer:
Eva8 [605]3 years ago
8 0

Answer: 29.17m/s^2

Explanation:

Given the following :

Velocity = 525 m/s

Time = 18 seconds

Acceleration = change in Velocity with time

Using the motion equation:

v = u + at

Where v = final Velocity

u = Initial Velocity and t = time

Plugging our values

525 = 0 + a × 18

525 = 18(a)

a = 525 / 18

a = 29.166666

a = 29.17 m/s^2

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78. A particle moves along the x- axis. The velocity of the particle at time tis given by 4vt()=3 t +1 . If the position of the
RoseWind [281]

Explanation:

Below is an attachment containing the solution

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3 years ago
8. When a 100 N bag of nails hangs motionless from a single vertical strand of rope, how many newtons of tension are exerted in
Svetllana [295]

If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
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If the bag is suspended from a single rope, then the tension
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3 years ago
Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

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r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

8 0
3 years ago
A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The
S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

\sum F_x = 0

f-N_w = 0

N_w = f

The forces in the horizontal direction would be,

\sum F_y = 0

N_f -W =0

N_f = W

The sum of Torques at equilibrium,

\sum \tau = 0

Wdcos\theta - N_wLsin\theta = 0

WdCos\theta = fLSin\theta

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The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

f_{max} = \mu W=\frac{Wd}{Ltan\theta}

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Replacing,

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Therefore the minimum angle that the person can reach is 46.9°

8 0
3 years ago
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