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
Darya [45]
2 years ago
10

Directed Reading for Section 2 - Acceleration

Physics
1 answer:
erastovalidia [21]2 years ago
5 0
Abs and piper have a great time and
You might be interested in
At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant accel
ratelena [41]
D.

50 mph - 30 mph= 20 mph net velocity
change.
20mph/3600 seconds/hour= .00555 MPS
.0055 miles per second


40 seconds to complete the change
.0055/40= .000138

7 0
2 years ago
100POINTTTTTSSS PLEASE HELP
attashe74 [19]
Components connected in series are connected along a single path, so the same current flows through all of the components. If the light bulbs are connected in parallel, the currents through the light bulbs combine to form the current in the battery, while the voltage drop is across each bulb and they all glow.
8 0
3 years ago
PLEASE HELP!!!<br>How are middle latitude cyclones and tropical cyclones different from each other?
defon
In contrast, extratropical cyclones have their strongest winds near the tropopause, which is about 8 miles above the surface. These differences are due to the tropical cyclone being “warm-core” in the troposphere, whereas extra-tropical cyclones are “warm-core” in the stratosphere and “cold-core” in the troposphere.
8 0
3 years ago
On a balanced seesaw, a boy three times as heavy as his partner sits
slega [8]

Answer:

1/3 the distance from the fulcrum

Explanation:

On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

W_1 d_1 = W_2 d_2

where

W1 is the weight of the boy

d1 is its distance from the fulcrum

W2 is the weight of his partner

d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

W_1 = 3 W_2

If we substitute this into the equation, we find:

(3 W_2) d_1 = W_2 d_2

and by simplifying:

3 d_1 = d_2\\d_1 = \frac{1}{3}d_2

which means that the boy sits at 1/3 the distance from the fulcrum.

8 0
3 years ago
Could a nucleus that has one proton but no neutrons exist?
diamong [38]
Yes, but there is only 1 atom like that and is is hydrogen. Hydrogen is the only element that could have a nucleus with one proton and no neutrons exist.
8 0
3 years ago
Other questions:
  • 1 what is global climate change. <br>2 consequences and the way forward​
    10·1 answer
  • If you want to lift a 5-kg box to a height of 2 meters, how much work must be
    5·2 answers
  • 1 2 3 4 5 6 7 8 9 10
    7·2 answers
  • A bowling ball collides with a pin, knocking it over. The ball continues to move
    6·1 answer
  • A boy with mass 50 kg running at 5m/s jumps on to a 20 kg trolley travelling in the same direction at 1.5 m/s. What is their com
    8·1 answer
  • Why does digital Radiology provide a reduced radiation dose compared to film?
    14·1 answer
  • The μ-receptor (mu) a. plays a role in analgesia and the rewarding effects of morphine. b. overlaps with the κ-receptor in its d
    11·1 answer
  • Pls help me with this one
    13·1 answer
  • How do you find the speed of a hraph
    9·1 answer
  • Ram and Ajay are trying to move a heavy box. Ram is applying 100N and Ajay is
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!