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3 years ago

Which line on the graph has the greatest slope?

Physics

2 answers:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

The black one

Explanation:

m stands for slope

Vadim26 [7]3 years ago

5 0

Answer:

the black line has the greatest slope

Explanation:

the black line has the greatest slope because if you at the green line its going right across the graph. the blue, yellow, and red lines arent the big of the slope as the black line so its the black line.

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The winch takes in cable at the constant rate of 130 mm/s. if the cylinder mass is 115 kg, determine the tension in cable 1. neg

nikitadnepr [17]

By applying Newton's second law of motion;

ma = mg - T

Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.

For the current case, the velocity is constant therefore,
a = 0

Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N

Tension in the cable is 1128.15 N.

8 0

3 years ago

Consider a collection of charges in a given region and suppose all other charges are distant and have a negligible effect. Furth

AlladinOne [14]

Answer:

E. Some charges in the region are positive, and some are negative.

Explanation:

Electric potential is given as;

$V = \frac{W}{Q}$

where;

W is the work done in moving a charge between two points which have a difference in potential

Q is quantity of charge in the given region

If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.

Therefore, the correct statement in the given options is "E"

E. Some charges in the region are positive, and some are negative.

3 0

3 years ago

I will run one mile in under 10 minutes 3 months from today.” is an example of a ___________ goal.

REY [17]

Answer:

long-term

Explanation:

usually a short term goal is able to be accomplished in a week or two. the question gives a 3 month time frame for the person to build up to the end goal.

4 0

2 years ago

A wave traveling at 230 m/sec has a wavelength of 2.1 meters. What is the frequency of this wave?

Galina-37 [17]

Answer:

Explanation:

The frequency of the wave when given it's velocity and wavelength only can be found by using the formula

$f = \frac{c}{ \lambda} \\$

where

c is the velocity of the wave in m/s

$\lambda$ is the wavelength in meters

From the question we have

c = 230 m/s

$\lambda$ = 2.1 m

$f = \frac{230}{2.1} = 109.523809... \\$

We have the final answer as

Hope this helps you

3 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago