A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?

What is a wave coming into a barrier

3241004551 [841]

A wave looses its power as it comes to shore because it gets less deeper every second it gets closer to shore

6 0

3 years ago

You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.

melomori [17]

Answer:

Explanation:

a)Magnitude = $\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }$

84= $\sqrt{(0- (-67))^{2} + (x-0)^{2} }$

x= +50.67 or -50.67 units

b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.

To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.

Magnitude = $\sqrt{(0- (67))^{2} + (-130.67)^{2} }$ = 146.85 units

c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - $Tan^{-1}(67/130.67)degrees$ i.e 152.85 degrees from the +ve x-axis.

5 0

3 years ago

This problem will require you to make some assumptions, so be sure that your assumptions are clear. You have dinner reservations

sertanlavr [38]

Answer:

No

Explanation:

The statement is implying that if you do a certain amount of work, you're entitled to do a certain amount of work. Sometimes, work doesn't entitle you a reward.

6 0

3 years ago

) you carry a 7.0 kg bag of groceries 1.2 m above the ground at constant velocity across a 2.7 m room. how much work do you do o

lapo4ka [179]

M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries

The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.

At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
= (7.0 kg)*(9.8 m/s²)*(1.2 m)
= 82.32 J

The change in PE is equal to the work done.

Answer: 82.3 J

3 0

3 years ago

A 2-kg wood block is pulled by a string across a rough horizontal floor. The string exerts a tension force of 30 N on the block

Luden [163]

Answer:

Work done, W = 84.57 Joules

Explanation:

It is given that,

Mass of the wooden block, m = 2 kg

Tension force acting on the string, F = 30 N

Angle made by the block with the horizontal, $\theta=20^{\circ}$

Distance covered by the block, d = 3 m

Let W is the work done by the tension force. It can be calculated as :

$W=F\ cos\theta\times d$

$W=30\times cos(20)\times 3$

W = 84.57 Joules

So, the work done by the tension force is 84.57 Joules. Hence, this is the required solution.

7 0

3 years ago