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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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All of the follow are ways of measuring volume execpt:
bogdanovich [222]
A). Scale.
You CAN measure volume with all the following except a scale.
6 0
3 years ago
A portable x-ray unit has a step-up transformer. The 120 V input is transformed to the 100 kV output needed by the x-ray tube. T
slega [8]

Answer:

 N_s\approx41667 \hspace{3}lo ops

Explanation:

In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:

\frac{V_p}{V_s} =\frac{N_p}{N_s} \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings

In this case:

V_p=120V\\V_s=100kV=100000V\\N_p=50

Therefore, using the previous equation and the data provided, let's solve for N_s :

N_s=\frac{N_p V_s}{V_p} =\frac{(50)(100000)}{120} =\frac{125000}{3} \approx41667\hspace{3}loo ps

Hence, the number of loops in the secondary is approximately 41667.

3 0
3 years ago
A car drives at a velocity of 6.8 m/s for 250 seconds. How far did it travel?
Leno4ka [110]

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the solution ~

As we know, Displacement =

\qquad \sf  \dashrightarrow \: velocity  \times time

\qquad \sf  \dashrightarrow \: 6.8 \times 250

\qquad \sf  \dashrightarrow \: 1700 \: m

<h3>OR </h3>

\qquad \sf  \dashrightarrow \: 1.7 \: km

The car travelled 1700 meters ( 1.7 km )

8 0
2 years ago
One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied
Alexandra [31]

Answer: 1.91*10^8 N/m²

Explanation:

Given

Radius of the steel, R = 10 mm = 0.01 m

Length of the steel, L = 80 cm = 0.8 m

Force applied on the steel, F = 60 kN

Stress on the rod, = ?

Area of the rod, A = πr²

A = 3.142 * 0.01²

A = 0.0003142

Stress = Force applied on the steel/Area of the steel

Stress = F/A

Stress = 60*10^3 / 0.0003142

Stress = 1.91*10^8 N/m²

From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²

8 0
2 years ago
| C₄H10<br> Number of hydrogen atoms?
Sunny_sXe [5.5K]

That's a molecule of Butane.

It's made out of 4 Carbon atoms and 10 Hydrogen atoms.

3 0
2 years ago
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