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3 years ago

6. Draw a velocity-time graph for an object originally traveling at -3 m/s. The object

Physics

1 answer:

faltersainse [42]3 years ago

4 0

See the graph in attachment

Explanation:

In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.

In the graph, we see that the initial velocity at time t = 0 is

$v_0 = -3 m/s$

and it is negative, so below the x-axis.

Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).

After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.

Learn more about velocity:

brainly.com/question/5248528

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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

3 years ago

PLS I NEED HELP ASAAP

Lerok [7]

Answer:

Option A

Explanation:

The Equation represents the displacement of the object which is represented by x

$x=t-t^2$

so, $x_0$ means when time is zero so we replace t with zero in the equation,

$x_0=(0)-(0)^2\\x_0=0$

now for v which is velocity we need to differentiate the function as the formula for velocity is rate of change of displacement over time so we derivate the equation once and get,

$v=1-2t\\$

now for $v_0$ we insert t = 0 and get

$v_0=1-2(0)\\v_0=1$

now for a which is acceleration the formula of acceleration is rate of change of velocity over time, so we differentiate the the equation of v(velocity) once or the equation of x(displacement) twice so now we get,

$a=-2$

so Option A is your answer.

Remember derivative of a constant is always zero because a constant value has no rate of change has its a constant hence the derivative is 0

5 0

2 years ago

What is your operational definition for a fast reaction time?

LuckyWell [14K]

Answer:

The ability to react to a certain stimulus with a speedy and effective manner

Explanation:

7 0

3 years ago

If you drop a silver dollar off a building and it hits the ground in 10 seconds, how fast was the coin going just before it hit?

Illusion [34]

Objects in free fall, disregarding terminal velocity, accelerate at 9.8(m/s)/s. so for every second it was falling, it gained 9.8m/s in speed. 9.8 * 10 = 98m/s

4 0

3 years ago

Read 2 more answers

A train brakes from 25 m/s to rest in 30 sec. What is its deceleration?

givi [52]

Answer:

a= -0.83m\s^2

Explanation:

a = v \ t

a = -25 \ 30 = -0.833 m\s^2

the object is slowing down 0.83 meter every second

8 0

3 years ago