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Paha777 [63]
3 years ago
8

A rope of negligible mass supports a block that weighs 30.0 N. as shown below. The maximum tension the rope can support without

breaking is 50.0 N. What is the maximum
acceleration the weight can be pulled upwards with without breaking the rope? (Use g = 10 m/s)
WON
Physics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

a = 6.7 m/s²

Explanation:

The formula to apply here is :

Mass * acceleration = Tension -weight

m*a = Ft -mg  -----where m is mass of block, a is acceleration , Ft  is force due to tension.

Given in the question that ;

Mass of block = 3 kg

Ft = 50 N

g= 10

m*a = Ft -mg

3*a = 50-30

3a= 20

a= 20/3

a= 6.7 m/s²

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Thepotemich [5.8K]

Answer:

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

Explanation:

From the image, we see that the resistors are connected in parallel. This means that the voltage passing through them is the same.

Now, formula for current is; I = V/R

In this case, V which is voltage is denoted by ε.

Thus;

I1 = ε/R1

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I3 = ε/R3

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3 years ago
Can you access an instance variable from a static method? explain why or why not.
marusya05 [52]
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7 0
2 years ago
A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-
Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

v_{s_{x}}=0

v_{s_{y}}=2.0\ m/s

The velocity of the boat in term x and y coordinate

For x component,

v_{b_{x}}=v_{b}\cos\theta

Put the value into the formula

v_{b_{x}}=5.60\cos19

v_{b_{x}}=5.29\ m/s

For y component,

v_{b_{y}}=v_{b}\sin\theta

Put the value into the formula

v_{b_{y}}=5.60\sin19

v_{b_{y}}=1.82\ m/s

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}

Put the value into the formula

v_{sb_{x}=0-5.29

v_{sb}_{x}=-5.29\ m/s

For y component,

v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}

Put the value into the formula

v_{sb_{x}=2.-1.82

v_{sb}_{x}=0.18\ m/s

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0
3 years ago
9. Ripening is a _______________ and spoiling is a ______________.
IrinaK [193]

Answer:

C

Explanation:

Chemical reaction,physical

6 0
2 years ago
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