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tatyana61 [14]
3 years ago
15

X(????) = 5.0???? 2 − 4.0???? 3 m.

Physics
1 answer:
Burka [1]3 years ago
4 0

Answer:

JA

Explanation:

s of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position. Assume all variable and constants are in SI units.

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It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration
bearhunter [10]

Answer:

About 12 seconds

Explanation:

6 0
3 years ago
2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th
Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = 5.33 × 10^{-5} T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ B = \frac{\mu_{0}i }{2\pi R}

Where B = magnetic field due to long wires, \mu_{0} = 4\pi \times10^{-7}, R = perpendicular distance from wire to given point

From any one wire R_{1}  = 5 cm, R_{2}  = 3 cm

so we write,

∴ B = B_{1} + B_{2}

 B = \frac{\mu_{0} i}{2\pi R_{1} } +  \frac{\mu_{0} i}{2\pi R_{2} }

 B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]

 B = 5.33\times10^{-5}  T

Therefore, the magnitude of magnetic field at given point = 5.33\times10^{-5} T

3 0
3 years ago
A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,
Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. <span>

An initial condition has been supplied by the problem: 

s=1.80m when t=0.245s 

<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2) 

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation): 

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2 

v=8.549m/s 

<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t) 

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like 

0=(v)+(a)(t) 

Substituting and solving for t: 

0=(8.549m/s)+(-9.81m/s^2)(t) 

t=0.8714s 

<span>Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m<span>

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window: 

3.725m – 1.8m=1.925m</span>

 

Answer:

<span>1.925m</span>

3 0
3 years ago
Which of the following forces is MOST involved in radioactive decay?
mario62 [17]
Weak nuclear force is the most involved
3 0
3 years ago
Read 2 more answers
If a planet has a radius 20% greater than that of the Earth but has the same mass as the Earth, what is the acceleration due to
Vaselesa [24]

Answer:

g_2=6.8125 m^2/sec

Explanation:

We know that

Acceleration due to gravity g is given by the formula

g= \frac{GM}{r^2}

G= gravitational constant

M= mass of the Earth

r= radius of the earth

g_1 = GM/r_1^2

Let acc. due to gravity after radius is 20% greater be g_2

then

g_2=GM/r_2^2

=> g1/g2 = (r_2/r_1)^2 => g2 = 9.81/1.2^2 = 6.8125

5 0
3 years ago
Read 2 more answers
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