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4 years ago

Can someone please here me I’m confused

Physics

1 answer:

Naddika [18.5K]4 years ago

5 0

I would say z, x, y, w

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Inside most ball-point pens is a small spring that compresses as the pen is pressed against the paper. If a force of 0.1 N compr

AnnZ [28]

Answer:

20 N/m

Explanation:

From the question,

The ball-point pen obays hook's law.

From hook's law,

F = ke............................ Equation 1

Where F = Force, k = spring constant, e = compression.

Make k the subject of the equation

k = F/e........................ Equation 2

Given: F = 0.1 N, e = 0.005 m.

Substitute these values into equation 2

k = 0.1/0.005

k = 20 N/m.

Hence the spring constant of the tiny spring is 20 N/m

8 0

3 years ago

a 4m long straight wir that carries acurrent of 0.5A is placed perpendicular to a uniform magnetic field. if the size of magneti

PolarNik [594]

Answer:

B=0.2T

Explanation:

given required solution

l=4m B=? F=BIL

i=0.5A B=F/IL

F=0.4N B=0.4N/0.5A*4m

B=0.4/2=0.2T

5 0

3 years ago

How far will an object move in 6 seconds if its average speed during that time is 45 m/s

babunello [35]

Answer:

270 Meters

Explanation:

45×6=270

3 0

3 years ago

Read 2 more answers

The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co

djyliett [7]

Answer:

11.06m/s²

Explanation:

According to Newtons second law of motion

$\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\$

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

Hence the the magnitude of the resulting acceleration is 11.06m/s²

6 0

3 years ago

A horizontal uniform meter stick supported at the 50-cm mark has a mass of 0.50 kg hanging from it at the 20-cm mark and a 0.30

ElenaW [278]

Answer:

70 cm

Explanation:

0.5 kg at 20 cm

0.3 kg at 60 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$0.5(50-20)=0.3(60-50)+0.6x\\\Rightarrow x=\dfrac{0.5(50-20)-0.3(60-50)}{0.6}\\\Rightarrow x=20\ cm$

The position of the third mass of 0.6 kg is at 20+50 = 70 cm

7 0

4 years ago

Read 2 more answers