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kirill115 [55]
3 years ago
13

A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s squared for 10.0 seconds. What is its velocity after the

acceleration????
Physics
1 answer:
zavuch27 [327]3 years ago
5 0
'  +4 m/s² ' means that the pigeon's speed is  4 m/s greater every second.

Starting from zero speed, after 10 seconds, its speed is 

                           (10 x 4m/s)  =  40 m/s.

We can't say anything about its velocity, because we have
no information regarding the direction of its flight.
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In a typical badminton swing the racket is in contact with the birdy for about 0.0010 seconds. If the 0.045kg birdy acquires a s
zepelin [54]

Answer:

3015 N

Explanation:

From Newton's second law, we know that;

F.t = mv

F = force on the ball= ?

m= mass of the ball

v= velocity

F= mv/t

F= 0.045 × 67/0.0010

F= 3.015/0.0010

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3 0
2 years ago
Physics help please
olchik [2.2K]

Answer:

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8 0
3 years ago
Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.
lukranit [14]

Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, q=3\times 10^{-9}\ C

Distance of the given point from the point charge is, d=3\ cm=0.03\ m

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

E=\frac{kq}{d^2}

Plug in k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m. Solve for 'E'.

E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

3 0
3 years ago
In a nuclear fusion reaction two 2H atoms are combined to produce one 4He.
Mrrafil [7]

Answer:

a)= 0.025602u

b) = 23.848MeV

c) N = 1.546 × 10¹³

Explanation:

The reaction is

²₁H   +   ²₁H   ⇄   ⁴₂H + Q

a) The mass difference is

Δm = 2m(²₁H) - m (⁴₂H)

       = 2(2.014102u) - 4.002602u

        = 0.025602u

b) Use the Einstein mass energy relation ship

The enegy  release is the mass difference times 931.5MeV/U

E = (0.025602) (931.5)

   = 23.848MeV

c)

the number of reaction need per seconds is

N = Q/E

     = 59W/ 23.848MeV

  = \frac{59}{(23.848 \times 10^6 )(1.6 \times 10^1^9) } \\\\= 1.546 \times 10^1^3

N = 1.546 × 10¹³

5 0
3 years ago
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viktelen [127]

Answer:

this is your solution

Explanation:

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