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3 years ago

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Dos masas están conectadas por una cuerda ligera que pasa por una polea sin rozamiento. Determine la aceleración de las masas y

la tensión de la cuerda si m Kg A  20 , m Kg B  50 y  0.20  K A Y B

1 answer:

Alexandra [31]3 years ago

5 0

Answer:A

Explanation:

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If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

torisob [31]

There's no such thing as "stationary in space". But if the distance
between the Earth and some stars is not changing, then (A) w<span>avelengths
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3 years ago

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

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3 years ago

What are three major types of fungi and an example of each.

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2) molds example is Rhizopus a type of mold that appears on old bread

3) mushrooms example is Amanita Phalloides also known as the "Death Cap " is a very poisonous mushroom and should not be ingested

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3 years ago

Read 2 more answers

A 26-kg sled is on a snow-covered slope. The coeﬃcients of friction between the sled’s runners and the snow are µs = 0.096 and µ

sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction $\mu _s=0.096$

coefficient of kinetic friction $\mu _k=0.072$

In order to move sled from rest we need to provide a force greater than static friction which is given by

$f_s=\mu mg=0.096\times 26\times 9.8=24.46 N$

After Moving Sled kinetic friction comes in to play which is less than static friction

$f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N$

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

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3 years ago

A mother is pulling a sled at constant velocity by means of a rope at 37°. The tension on the rope is 120 N. Mass of children pl

nordsb [41]

Answer:

Please find the complete solution in the attached file.

Explanation:

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3 years ago