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topjm [15]
2 years ago
12

A FORCE OF 3,200 N WEST is applied to a 160 kg mass. What is the acceleration of the mass

Physics
1 answer:
Luba_88 [7]2 years ago
3 0
30098
. Hi I’m sorry to hear about the family and stuff but I’m still coming in for the day so I’m not going out to the beach today or just a little later lol lol I’m going on the beach with the girls lol I just need a shower lol I need a shower lol I’m just going on the
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A plane glides towards a ground-based radar dish. Radar locates the plane at a distance D = 22 km from the dish, at an angle θ =
vfiekz [6]

The correct answer is :

Unit vectors I and j along the x-axis and y-axis, respectively, define the Cartesian coordinate system. The radial unit vector r, which indicates the direction from the origin, and the unit vector t, which is orthogonal (perpendicular) to the radial direction, together create the polar coordinate system.

We can obtain the horizontal component by applying the trigonometric identity of Cos(Ф), and if we obtain the component on the x axle, such as 22000 (m)×Cos(51°) = x, we may determine that x = 13845.05 metres. We need to obtain the vector components because we already know the distance and the angle.

To learn more about Cartesian unit-vector refer the link:

brainly.com/question/26776558

#SPJ9

8 0
1 year ago
How far from the base of the platform does she land?
seropon [69]

When Janet leaves the platform, she's moving horizontally at 1.92 m/s.  We assume that there's no air resistance, and gravity has no effect on horizontal motion.  There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.

She's in the air for 1.1 second before she hits the water.

Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform

(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>

3 0
3 years ago
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it, as shown in the figure above.
cestrela7 [59]

Explanation:

Forgive me If I am wrong, it's been a while since I've studied Torque.

The formula for the angular momentum is

momentum= I*w.

We can also write I*W as 1/2MR^2 * W so the extra mass coming from the block of clay would most likely cause the angular momentum to increase from the amount it was before.

6 0
2 years ago
A water-balloon launcher with a mass of 1.75 kg is suspended on a wire. It fires a 1.25 kg balloon to the east at a velocity of
poizon [28]

Answer:

8.6 to west

Explanation:

8 0
3 years ago
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