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Lady_Fox [76]
3 years ago
12

A cat chases a mouse until the cat gets tired. A graph of the cat's velocity over time is shown below. Velocity 8+ 6+ 9+ 2 Time

is) 2 3 What is the cat's average acceleration from 12 to 24 seconds? Answer using two digrii cont ligures​
Physics
2 answers:
iragen [17]3 years ago
8 0

Answer:

-0.67

Explanation:

khan academy

(final velocity minus initial velocity) divided by time

defon3 years ago
6 0

Answer:

-0.67

Explanation:

khan academy

You might be interested in
A parallel beam of light of wavelength 4.5 x 10^-7 m is incident on a pair of slits that are 5.0 x 10^-4 m apart. The interferen
RSB [31]

Answer:

1.8x10⁻³m

Explanation:

From the question above, the following information was used to solve the problem.

wavelength λ = 4.5x10⁻⁷m

Length L = 2.0 meters

distance d = 5 x 10₋⁴m

ΔY = λL/d

= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m

= 0.00000045 / 0.0005

= 0.0000009/0.0005

= 0.0018

= 1.8x10⁻³m

from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m

thank you!

3 0
3 years ago
An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
3 years ago
How far does a car move when it travels 50 miles an hour for 4 hours?
Alex_Xolod [135]

Answer:

200 mph

Explanation:

50 times 4 equals 200

6 0
3 years ago
special metal hangers or stirrips called joist hangers are used when joist must be _____ the bottom of the grider or beam
djyliett [7]

Answer:

Special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.

Explanation:

  • A joist hanger also known as a beam hanger is a mechanical device which is used to fasten joists and rafters.
  • The rafters are the carried members to beams and headers are the carrying members.

Thus, special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.

Learn more about construction here:

brainly.com/question/14428327

#SPJ4

7 0
2 years ago
Please answer for 20 points
Nat2105 [25]
C. Reduce friction between its moving parts
4 0
3 years ago
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