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3 years ago

This is a ______________________reaction.

Physics

2 answers:

Mars2501 [29]3 years ago

8 0

Answer: Chemical reaction?

Explanation:

i do not know it can be anything!

miskamm [114]3 years ago

7 0

Answer:

Chain reaction?

Explanation:

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The graph is tied to the reading

AleksandrR [38]

I. Positive acceleration increases velocity. Negative acceleration decreases velocity. runner A sped up until the finish line and then slowed to a stop.

ii. Zero a acceleration implies a constant, unchanging velocity not a zero velocity. runner B achieved some velocity prior to 8s and is moving and must slow down to reach a stop.

iii. None. No aspects of this reasoning are correct. Everything she says is wrong. See iv for what/why.

iv. The sign on acceleration denotes the direction of *change in velocity* not change in direction. The sign on velocity can denote change in direction but only “forward” or “reverse” along a particular path. Cardinal direction is not indicated, generally, by the sign on velocity. It may correspond to North/South situationally but it is not an built-in feature of velocity and its sign. For example, if you are traveling with positive velocity and turn left to continue your journey you still have a positive velocity in the new direction. In fact, if you turn left again, traveling in the opposite direction as the one you started with your velocity would still be positive… in the new direction. The velocity relative to original direction could be said to be negative but that would be a confusing way to describe a journey. Maybe if you stopped the vehicle and moved in reverse, you could meaningfully say velocity was negative.

5 0

3 years ago

A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N i

DaniilM [7]

Answer:

The apparent weight of the person as she pass the highest point is $N = 458.8 \ N$

Explanation:

From the question we are told that

The radius of the Ferris wheel is $r = 5.0 \ m$

The period of revolution is $T = 8.0 \ s$

The weight of the person is $W = 670 \ N$

Generally the speed of the wheel is mathematically represented as

$v = \frac{2 \pi r}{T }$

substituting values

$v = \frac{2 * 3.142 * 5}{8 }$

$v = 3.9 3 \ m/s$

The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as

$N = mg - \frac{mv^2}{r}$

Where m is the mass of the person which is mathematically evaluated as

$m = \frac{W}{g}$

substituting values

$m = \frac{670}{9.8}$

$m = 68.37 \ kg$

$N = 68.37 * 9.8 - \frac{68.37 * {3.93}^2}{5}$

$N = 458.8 \ N$

5 0

3 years ago

suppose that the man pictured on the front side is orbiting the earth (mass=5.98 x 10^24kg at a distance of 310 miles above the

wlad13 [49]

Answer:

a = 9.81[m/s^2]; v = 18683.5[m/s]

Explanation:

The stament of the problem is:

Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).

a. What acceleration does he experience due to the earth's pull?

b. What tangential velocity must he possess in order that he orbit safely (in m/s)?

First we need to convert all the initial data to units of the SI

Rs = 310 [mil] = 498897 [m]

RT= 400 [mil] = 643738 [m]

r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"

$F=G*\frac{M*m}{r^{2} } \\where:\\$

G = universal gravitational constant

M = mass of the earth [kg]

m = mass of the man [kg]

r = distance from the center of the earth to the man [m]

The acceleration he is experimenting is the same acceleration given by the gravity, therefore:

a = g = 9.81[m/s^2]

To find the tangential velocity, we must determinate the force exerted by the earth.

Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.

$G = 6.67*10^{-11} [\frac{N*m^{2} }{kg^{2} } ]\\M=5.98*10^{24}[kg]\\ m=100 [kg]\\Replacing:\\F = G*\frac{M*m}{r^{2} }$

$F = 6.67*10^{-11}*\frac{5.98*10^{24}*100 }{1142635^{2} } \\ F= 30550 [N]$

And this force will be equal to the following expression:

$F = m*\frac{v^{2} }{r} \\where:\\v= tangential velocity [m/s]\\v=\sqrt{\frac{F*r}{m} } \\v=\sqrt{\frac{30550*1142635}{100} } \\v=18683.5[m/s]$

8 0

3 years ago

FOR FLVS STUDENTS ONLY///NO BIT LINKS, SCAM LINKS, OR GIBBERSH

Marysya12 [62]

Answer:

whats thw question tho

Explanation:

6 0

3 years ago

A uniform, 4.5 kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its

Akimi4 [234]

Answer:

a) Angular speed(w) = 2.02rad/sec

b) 73J ( It is Inelastic Collision)

Explanation:

Given:

Mass=45kg

Length on each side = 1.5m side which is hangs vertically from a frictionless pivot at the center of its upper edge.

We need to calculate

(a) What is the angular speed and

(b) To know why the angular momentum conserved but not the linear momentum

CHECK THE ATTACHMENT FOR DETAILED EXPLATION

6 0

3 years ago