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3 years ago

6

Find the tension in the two groups that are holding the 2.9 kg object in Pl., One makes an angle of 35.6° with respect to the ve

rtical group 2 is pulling horizontally

1 answer:

Stella [2.4K]3 years ago

4 0

Answer:

≈ 20.35 N [newton's of tension]

Explanation:

( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =

( (28.42) ÷ (≈0.813) ) × (≈0.582) =

(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35

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A student pushes a 10 kg box with a force of 30 N at an angle 45 degrees below the horizontal . The box experi 21.2 N of force a

Shkiper50 [21]

Answer:

Acceleration =0

Explanation:

There are two forces on the box in horizontal direction.

First one is horizontal component of 30N at 45°.

$F=\frac{1}{\sqrt{2} }*30=21.2$

Another one is given in opposite to direction of motion mean opposite to this force.

Which is = 2.22 N

Now we know F (net) =mass*acceleration(center of mass)

21.2-21.2 =10*acceleration

acceleration =0

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4 years ago

We observe a glowing cloud of gas in space with a spectroscope. We note that many of the familiar lines of hydrogen that we know

konstantin123 [22]

Answer:

No information can be concluded from this observation alone the cloud is much cooler than hydrogen on Earth

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3 years ago

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.45 m/s^2 for 20.0 s.

Anton [14]

Answer:

a) The total displacement of the trip was 5.32 × 10³ m

b) The average speeds were:

leg 1: 24.5 m/s

leg 2: 49 m/s

leg 3: 23.9 m/s

Complete trip: 43.8 m/s

Explanation:

The position and velocity equations for an object moving along a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

If the velocity is constant, then a = 0 and x = x0 + v · t where "v" is the velocity.

a) To calculate the total displacement of the trip, let´s calculate the distance traveled in each phase.

Phase 1:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 0 m/s · t + 1/2 · 2.45 m/s² · (20.0 s)²

x = 490 m

The velocity reached in that phase is:

v = v0 + a · t

v = 0 m/s + 2.45 m/s² · 20.0 s

v = 49.0 m/s

Phase 2:

x = x0 + v · t

x = 490 m + 49.0 m/s · 96.0 s

x = 5.19 × 10³ m

Phase 3:

x = x0 + v0 · t + 1/2 · a · t²

x = 5.19 × 10³ m + 49 m/s · 5.44 s - 1/2 · 9.00 m/s² · (5.44 s)²

x = 5.32 × 10³ m

The total displacement of the trip was 5.32 × 10³ m

b) The average speed is calculated as the traveled distance divided by the elapsed time:

average speed v = final position - initial position / (final time- initial time)

Phase 1:

v = 490 m - 0 m / 20.0 s = 24.5 m/s

Phase 2:

v = 5.19 × 10³ m - 490 / 96.0 s

v = 48.9 m/s (without rounding the final position the result is 49.0 m/s)

Phase 3:

v = 5.32 × 10³ m - 5.19 × 10³ m / 5.44 s = 23.9 m/s

For the complete trip:

v = 5.32 × 10³ m - 0 m / (20.0 s + 96.0 s + 5.44 s)

v = 43.8 m/s

7 0

4 years ago

In what sense is the earth a closed system?

I am Lyosha [343]

Answer:D

Explanation: just D.)

4 0

3 years ago

Read 2 more answers

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total di

S_A_V [24]

Answer:

1keff=1k1+1k2

see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.

Now the spring is in series combination

F $\alpha$ e

F=ke

k=f/e.........*

where k is the force constant or the constant of proportionality

k=f/e

$f_{eff} =f_{1} +f_{2}$ ............................1

also for effective force constant

divide all through by extension

1) Total force is

Ft=F1+F2

Ft=k1e1+k2e2

F = k(e1+e2) 2)

Since force on the 2 springs is the same, so

k1e1=k2e2

e1=F/k1 and e2=F/k2,

and e1+e2=F/keq

Substituting e1 and e2, you get

1/keq=1/k1+1/k2

Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.

4 0

3 years ago