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ludmilkaskok [199]

3 years ago

11

A 2200-kg auto moving northward at 12.0 m/s runs into a 3800-kg truck which is also moving northward, but at 5.00 m/s. If the ve

hicles lock bumpers, how fast are they moving just after the collision

1 answer:

Nookie1986 [14]3 years ago

8 0

Answer:

7.56 m/s

Explanation:

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Which BEST describes the difference between speed and velocity?

Murljashka [212]

D. velocity includes rate of change and direction

8 0

3 years ago

Read 2 more answers

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

4 years ago

A car starts from rest and accelerates at 5 m/s/s.

goldenfox [79]

Please find attached photograph for your answer. Hope it helps. Please do comment

3 0

3 years ago

What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a

Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

$A = W + \dfrac{W}{2}$

$A =\dfrac{3W}{2}$

$A =\dfrac{3mg}{2}$

When the car is at the bottom, the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

$N = m g + \dfrac{mv^2}{r}$

$\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}$

$\dfrac{mg}{2} = \dfrac{mv^2}{r}$

$v = \sqrt{\dfrac{rg}{2}}$

$v = \sqrt{\dfrac{40\times 9.8}{2}}$

v = 14 m/s

8 0

3 years ago

Find the cost of excavating a space 84 ft long, 42 ft wide, and 9 ft deep at a cost of $39/yd3. (simplify your answer completely

m_a_m_a [10]

The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864

Information about the problem:

Space long= 84 ft
Space wide= 42 ft
Space deep= 9 ft
Cost by yard3 = $39/yd3
Total cost= ?

To solve this problem, we have to state the equation using the information of the problem:

Calculating the volume of the total space:

space volume = space long * space wide * space deep

space volume = 84 ft * 42 ft * 9 ft

space volume = 31752 ft3

By converting the volume from ft3 to yd3, we have:

31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3

Calculating the cost of excavating the volume space:

Total cost = space volume * cost by yard3

Total cost = 1176 yd3 * $39/yd3

Total cost = $45864

<h3>What is volume?</h3>

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

Learn more about volume at: brainly.com/question/12628341

#SPJ4

4 0

1 year ago