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Gelneren [198K]

3 years ago

9

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?

1 answer:

anygoal [31]3 years ago

3 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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The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

$F=\frac{Gm_{1} m_{2} }{d^{2} }$ ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

$1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }$

$d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}$

$d=\sqrt{1.48\times10^{17}}$

d = 3.85 x 10⁸ m

3 0

3 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago

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earnstyle [38]

The answer is A. ive done a 5-k race, so its for sure 3 miles.

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3 years ago

If the net external force acting on a system is zero , then the total momentum of the system is zero

Neporo4naja [7]

If net external force acting on the system is zero, momentum is conserved. That means, initial and final momentum are same → total momentum of the system is zero.

8 0

3 years ago

Read 2 more answers

A cat jumps off a piano that is 1.3m high. The initial velocity of the cat is 3m/s at an angle of 37degrees above the horizontal

SpyIntel [72]

Answer:

$x=1.75m$

Explanation:

From the exercise we have that

$y_{o}=1.3m\\v_{o}=3m/s, \beta =37\\$

<em><u>To find how far from the edge of the piano does the cat strike the floor, we need to calculate its time first </u></em>

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

At the end of the motion y=0m

$0=1.3+3sin(37)t-\frac{1}{2}(9.8)t^{2}$

Solving for t

$t=-0.36 s$ or $t=0.73s$

Since the <u>time</u> can't be negative the answer is t=0.73

Knowing that we can calculate how far does the cat strike the floor

$x=v_{ox}t=3cos(37)(0.73)=1.75m$

6 0

3 years ago