Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
Explanation:
From the given information:
the car's momentum = momentum of the truck
∴
(a) 816 kg × v = 2650 kg × 16.0 km/h
v = (2650 kg × 16.0 km/h) / 816 kg
v = 51.96 km/hr
(b) 816 kg × v = 9080 kg × 16.0 km/h
v = (9080 kg × 16.0 km/h) / 816 kg
v = 178.04 km/hr
Choose the CORRECT statements. The superposition of two waves.
I. refers to the effects of waves at great distances.
Il. refers to how displacements of the two waves add together.
Ill. results into constructive interference and destructive interference
IV. results into minimum amplitude when crest meets trough.
V. results into destructive interference and the waves stop propagating.
A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V
Answer:
A
Explanation:
I guess not that much confidential!
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
To solve more questions on Work, Energy and Power, visit the link below -
https://brainly.com/question/208670
#SPJ2
A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component of the force?
A) 100N
B) 196N
C) 50N
D) 86N
show your work please
Answer:
the horizontal component of the force is 50 N
Explanation:
Given;
force applied by the man, F = 100 N
angle of inclination of the force, θ = 60⁰
mass of the dog, m = 20 kg
The horizontal component of the force is calculated as;
[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]
Therefore, the horizontal component of the force is 50 N
The wavelength of visible light range of 400 to 750mm .what is the corresponding range of photon energies for visible light
Answer:
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Explanation:
The energy of a photon is calculated using the following equation;
E = hf
where;
h is Planck's constant = 6.63 x 10⁻³⁴ Js
f is frequency of the photon
[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]
[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?
