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2 years ago

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A 0,9 -kg object attached to the end of a string swings in a vertical circle (radius = 75 cm). At the top of the circle the spee

d of the object is 6,5 m/s. What is the magnitude of the tension in the string at this position?

1 answer:

omeli [17]2 years ago

6 0

0.6 cm is the answer add it up and find the m/s hope this helps

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A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 100 N is being applied to the sled rope

tangare [24]

Answer:

60.18 N

Explanation:

Given that:

The force applied on the sled = 100 N

Suppose, the angle between the sled rope and the ground = 53°

The horizontal force which acts in the horizontal direction can be expressed as:

$F_x = F \ cos \theta$

$F_x = 100 \ cos (53)$

$F_x = 60.18 \ N$

But if the angle between the sled rope is parallel to the ground. Then, we use an angle on a straight line which is = 180°

$F_x = F \ cos \theta$

$F_x = 100 \ cos (180)$

= 100 × -1

= -100 N

3 0

3 years ago

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

I am Lyosha [343]

Answer:

$2.09\ \text{m/s}$

$22298.4\ \text{J}$

Explanation:

m = Mass of each the cars = $1.6\times 10^4\ \text{kg}$

$u_1$ = Initial velocity of first car = 3.46 m/s

$u_2$ = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

$mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}$

Speed of the three coupled cars after the collision is $2.09\ \text{m/s}$ .

As energy in the system is conserved we have

$K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}$

The kinetic energy lost during the collision is $22298.4\ \text{J}$ .

6 0

3 years ago

I don't understand this question at all, can I please get some help?

beks73 [17]

V^2/R=180W
v=root 180R

4 0

3 years ago

A wildebeest and chicken participate in a race over a 2.00km long course. the wildebeest travels at a speed of 16.0m/s and chick

Nezavi [6.7K]

Answer:

(a) The distance of the chicken from the finish line is 62.5 m

(b) The stationary time of the wildebeest is 675 s

Explanation:

Given;

total distance traveled by wildebeest and chicken, d = 2 km = 2000 m

speed of the wildebeest, $v_w$ = 16 m/s

speed of the chicken, $v_c$ = 2.5 m/s

Time for wildebeest to finish the race without stopping, 2000 / 16 = 125 s

Time for chicken to finish the race without stopping, 2000/2.5 = 800 s

(b) for how long in time (in s) was the wildebeest stationary?

t(stationary) = t(chicken) - t(wildebeest)

t = 800s - 125 s

t = 675 s

(a) how far (in m) is the chicken from the Finish Line when the wildebeest resume the race?

The time taken for the wildebeest to run 1.6 km (1600 m) is given by;

t = 1600 / 16 = 100 s

The total time spent by the wildebeest before it resumed the race = stationary time + 100s

t (total) = 675 s + 100 s = 775 s

Distnace traveled by the chicken when the wildebeest resumed the race = 2.5m/s x 775s = 1937.5 m

Thus, the distance of the chicken from the finish line = 2000 m - 1937.5 m

the distance of the chicken from the finish line = 62.5 m

7 0

3 years ago

HELP ME PLEASE!!!!!! A table showing variety among some young people.

Finger [1]

I’m not sure if you want some ideas but like if you go then eye colour hair colour and then genders?

4 0

3 years ago