What variables should you keep constant in initial speed

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo

denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

$\mu _smg=610$

$\mu _s\times 93\times 9.8=610$

$\mu _s=0.669$

5 0

3 years ago

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal

kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;

A: vA (t) = ť^2 – 5t + 20

B: vB (t) = t^2+ 3t + 10

Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

vA = 20 m/s

For vB

vB (0) = 0^2+ 3(0) + 10

vB = 10 m/s

We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

3 0

3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$