1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
laiz [17]
2 years ago
14

Science questions!! Please help!!

Physics
1 answer:
kondor19780726 [428]2 years ago
7 0

Answer:

1. b

2. c

3. b

4. a

5. a

6. c

7. c

8. d

You might be interested in
Show that any three linear operators A, B, and Ĉ satisfy the following (Ja- cobi) identity (10 pt) [[A, B] Ĉ] + [[B,C), A] + [[C
DerKrebs [107]

Answer:

Three linear operators A,B, and C will satisfy the condition [[A, B],C] + [[B,C), A] + [[C, A], B] = 0.

Explanation:

According to the question we have to prove.

[[A, B],C] + [[B,C), A] + [[C, A], B] = 0

Now taking Left hand side of the equation and solve.

[[A, B],C] + [[B,C), A] + [[C, A], B]

Now use commutator property on it as,

=[A,B] C-C[A,B]+[B,C]A-A[B,C]+[C,A]B-B[C,A]\\=(AB-BA)C-C(AB-BA)+(BC-CA)A-A(BC-CB)+(CA-AC)B-B(CA-AC)\\=ABC-BAC-CAB+CBA+BCA-CAB-ABC+ACB+CAB-ACB-BCA+BAC\\=0

Therefore, it is proved that [[A, B],C] + [[B,C), A] + [[C, A], B] = 0.

7 0
3 years ago
If a force of 40N is applied for 0.2 sec to change the momentum of a volleyball, what is the impulse?
Montano1993 [528]

Answer:

8ns

Explanation:

40(0.2)

=8ns

7 0
2 years ago
During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

5 0
3 years ago
Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea
Jet001 [13]

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon (m_{E}, m_{M}) are 5.972\times 10^{24}\,kg and 7.349\times 10^{22}\,kg, respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}

If x_{E} = 0\,km and x_{M} = 384,403\,km, then:

\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}

\bar x = 4.673\,km

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

8 0
2 years ago
A 23.5 kg object travels a distance of 85 m in 30 s. Find the momentum. (Hint: you must find speed first)
VLD [36.1K]
Yup, I think you add all of them
7 0
2 years ago
Other questions:
  • Which describes nuclear fusion?
    11·2 answers
  • Sully is riding a snowmobile on a flat, snow-covered surface with a constant velocity of 10 meters/second. The total mass of the
    10·2 answers
  • Is the eccentricity of planet orbits closer to 1 or 0?
    10·1 answer
  • A person stands on a scale in an elevator. his apparent weight will be the greatest when the elevator:
    8·2 answers
  • Select the correct answer. Five marbles roll down a ramp. Each marble reaches the bottom of the ramp at a speed of 3 meters/seco
    9·2 answers
  • A car of mass 300kg starts from rest and travels upwards along a straight road of 450m inclined at an angle of 5 degree to the h
    5·1 answer
  • Suppose you are driving a car around in a circle of radius 212 ft, at a velocity which has the constant magnitude of 43 ft/s. A
    9·1 answer
  • I WILL MARK BRAINLIST!!!!
    10·1 answer
  • How are volt and electron volt related?<br> How do they differ?
    10·1 answer
  • Why sonic the hedgehog its so broken in power?​
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!