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astraxan [27]
3 years ago
13

Two horses on opposite sides of a narrow stream are pulling a barge up the stream. Each hors pulls with a force of 720N. The rop

es from the horses meet at a common point on the front of he barge. What is the resultant force exerted by the two horses if the angle between the rope is 60 degrees?
Physics
1 answer:
Sonja [21]3 years ago
8 0
For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N
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which of the following laboratory tool would be most appropiate for measuring the approximate volume of a liquid?
SIZIF [17.4K]
Beaker would be most appropriate for measuring the approximate volume of a liquid.
7 0
3 years ago
A 1.0 kg rock is thrown straight upward with an initial speed of 8.0 m/s. What is its speed
Ronch [10]

Answer:5.7m/s

Explanation:

Mass=1kg

Initial velocity=u=8m/s

height=h=1.6m

Final velocity =v

Acceleration due to gravity=g=9.8m/s^2

v^2=u^2-2xgxh

v^2=8^2-2x9.8x1.6

v^2=8x8-2x9.8x1.6

v^2=64-31.36

v^2=32.64

Take the square root of both sides

√(v^2)=√(32.64)

v=5.7

Speed at the height of 1.6m is 5.7m/s

8 0
3 years ago
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

6 0
3 years ago
What do parallel circuits and series circuits have in common
Sonbull [250]
Voltage of each component is same.
7 0
3 years ago
A series RCL circuit contains a 65.2-Ω resistor, a 2.26-μF capacitor, and a 2.08-mH inductor. When the frequency is 2400 Hz, w
raketka [301]

Answer:

The power factor of the circuit is 0.99

Explanation:

Given;

resistance of the resistor, R = 65.2 ohms

capacitance of the capacitor, C = 2.26 μF = 2.26 x 10⁻⁶ F

inductance, L = 2.08 mH = 2.08 x 10⁻³ H

frequency of the AC, f = 2400 Hz

The angular frequency is given by;

ω = 2πf

ω = 2π(2400) = 15081.6 rad/s

The inductive reactance is given by;

XL = ωL

XL = (15081.6 x 2.08 x 10⁻³)

XL = 31.37 ohms

The capacitive reactance is given by;

X_c = \frac{1}{\omega C} \\\\X_c = \frac{1}{(15081.6)(2.26*10^{-6} )}\\\\X_c = 29.34 \ ohms

The phase difference is given by;

tan\phi = \frac{X_l - X_c}{R}\\\\ tan\phi =\frac{31.37-29.34}{65.2} \\\\tan\phi = 0.0311 \\\\\phi  = tan^{-1} (0.0311)\\\\\phi  = 1.78^0

The power factor is given by;

CosФ = Cos(1.78) = 0.99

Therefore, the power factor of the circuit is 0.99

5 0
3 years ago
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