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vladimir2022 [97]
3 years ago
9

What is the action force of a flying bird?

Physics
1 answer:
kirill [66]3 years ago
7 0
It is called the reaction force of a bird flying.
You might be interested in
Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N
Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0
1 year ago
In DC motor, the split rings are made of <br> A) steel B) copper C) wood D) glass
USPshnik [31]

I think is

(B) Copper

8 0
3 years ago
Describe what happens to chemical<br> bonds during a chemical reaction
dimulka [17.4K]

Explanation:

The bonds that keep molecules together break apart and form new bonds during chemical reactions, rearranging atoms into different substances. Each bond takes a distinct amount of energy to either break or form; the reaction does not take place without this energy, and the reactants stay as they were.

6 0
2 years ago
Your oven has a power rating of 5000 watts. How many kilowatts is this ?
Gekata [30.6K]

1,000 watts = 1 kilowatt
2,000 watts = 2 kilowatts
3,000 watts = 3 kilowatts
4,000 watts = 4 kilowatts
<em>5,000 watts = 5 kilowatts</em>

8 0
3 years ago
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0
3 years ago
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