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skad [1K]
3 years ago
6

A stone is thrown straight up. when it reaches it’s highest point, _____

Physics
1 answer:
Sati [7]3 years ago
8 0

both it velocity and acceleration is zero.

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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,
valina [46]

Answer:

=2.99\times10^9J

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement S = 0.72\times10^3m

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19°   ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)

=2.99\times10^9J

6 0
3 years ago
The basis for defining the length of a day is the fact that question 21 options:
abruzzese [7]
B.earth rotates on its axis in 24 hours; i.e., it rotates 15° of longitude per hour.
6 0
3 years ago
During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between
Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0
3 years ago
Read 2 more answers
A ball is kicked at a speed of 16m/s at 33° and it eventually returns to ground level further down field.
saw5 [17]

Hi there!

We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).

Remember to break the velocity into its vertical and horizontal components.

Thus:

0 = vi - at

0 = 16sin(33°) - 9.8(t)

9.8t = 16sin(33°)

t = .889 sec

Find the max height by plugging this time into the equation:

Δd = vit + 1/2at²

Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²

Solve:

Δd = 7.747 - 3.873 = 3.8744 m

4 0
3 years ago
Read 2 more answers
With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t
VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu  = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v =  \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N

Therefore, the tension of the rope is 34.95 N

4 0
3 years ago
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