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Lera25 [3.4K]
3 years ago
12

8. An object is moving with a constant deceleration of 6 m/s. If its initial speed is

Physics
1 answer:
UNO [17]3 years ago
8 0

Explanation:

we use the formula, v=u+at

v=30+(-6)2=18 m/s

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Which best describes a nonmagnetic material?
Natali [406]
Material that is not attracted to metal

3 0
3 years ago
Read 2 more answers
To construct a solenoid, you wrap insulated wire uniformly around a plastic tube 7.1 cm in diameter and 57 cm in length. You wou
ASHA 777 [7]

Answer:

We need about 8769 meters of wire to produce a 2.6 kilogauss magnetic field.

Explanation:

Recall the formula for the magnetic field produced by a solenoid of length L. N turns, and running a current I:

B=\mu_0\,\frac{N}{L} \,I

So, in our case, where B = 2.6 KG = 0.26 Tesla; I is 3 amperes, and L = 0.57 m, we can find what is the number of turns needed;

B=\mu_0\,\frac{N}{L} \,I\\0.26=4\,\pi\,10^{-7}\frac{N}{0.57} \,3\\N=\frac{0.26*0.57\,10^7}{12\,\pi} \\N=39311.27

Therefore we need about 39312 turns of wire. Considering that each turn must have a length of \pi\,D, where D is the diameter of the plastic cylindrical tube, then the total length of the wire must be:

Length=39312\,(\pi\,D)=39312\,(\pi\,0.071)\approx 8768.66\,\,m

We can round it to about 8769 meters.

5 0
3 years ago
How much time will it take to do 33j of work with 11 w of power
Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s

 

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}

8 0
3 years ago
Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s
anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of -5.2^{\circ}

D) The impulse exerted on C is 4.29 kg m/s at a direction of -5.2^{\circ}

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B

m_A = 0.600 kg is the mass of sphere A

u_A = 4.00 m/s is the initial velocity of the sphere A (taking the right as positive direction)

v_A is the final velocity of sphere A

m_B = 1.80 kg is the mass of sphere B

u_B = 2.00 m/s is the initial velocity of the sphere B

v_B = 3.00 m/s is the final velocity of the sphere B

Solving for vA:

v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J

After the collision:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s

While the total momentum along the y-axis is zero:

p_y = 0

We can now write the equations of conservation of momentum along the two directions as follows:

p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy} (1)

We also know the components of the momentum of B after the collision:

p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s

So substituting into (1), we find the components of the momentum of C after the collision:

p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s

So the magnitude of the momentum of C is

p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s

And the direction is

\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

p_C = 0

While the final momentum is:

p'_C = 4.29 kg m/s

So the magnitude of the impulse exerted on C is

I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore -5.2^{\circ}.

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

F=Ma

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

#LearnwithBrainly

8 0
3 years ago
Which method involves using tools to control invasive species?
vitfil [10]

Answer;

Mechanical control

Explanation;

-Invasive species may cause environmental harm, economic harm, or impact human health. A key factor that makes many species invasive is a lack of predators in the new environment.

-Mechanical control usually refers to the mowing or mechanical cutting of an invasive plant infestation to limit seed production. With mowing, timing is essential. Invasive plants must be removed before the plants go to seed in order to be an effective method of control. Plants should be cut as close to the ground as possible and may have to be treated more than once in a growing season to achieve desired results.

5 0
3 years ago
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