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4 years ago

How much force is required to accelerate a 50 kg mass at 2 m/s2?

1 answer:

4 0

=0.5 x40
= (calculate it) 
2. F= -.45 x 41 
= 
3. F=20 x 12 
= 
4. F=50 x 30 
= 

I believe you can go on from here, remember its easy 

for the last 2 just rearange the formual 
for 9 it will be a=F/m and 10. m=f/a

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A black lift of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35

horsena [70]

Answer:

Explanation:

refractive index of ember = sin of angle of incidence / sin of angle of refraction

= sin 35 / sin24

= .5735 / .4067

= 1.41

This is refractive index of ember with respect to water

refractive index of ember with respect to water

= wμe = μe / μw

μe = wμe x μw

= 1.33 x 1.41

= 1.87

refractive index of ember with respect to air = 1.87 .

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3 years ago

What helped provide support for the Plate Tectonic theory in the 1960s?

skad [1K]

D. Mineral and fossil matches from tests done on different continents.

3 0

3 years ago

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Một nười thả 420g chì ở nhiệt độ 100°C và 260g nước ở nhiệt độ 58°C làm cho nước nóng lên tới 60°C . Cho nhiệt dung riêng của nư

vlabodo [156]

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6 0

3 years ago

A student wearing a frictionless roller skates on a horizontal is pushed by a friend with a constant force of 55N. How far must

umka21 [38]

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

362 = 55d

d = 362/55

d = 6.58m

Hence the student must push at a distance of 6.58m

3 0

3 years ago

A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no apprecia

masha68 [24]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

$V(t) = \epsilon. e^{-t.\frac{L}{R} }$

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

$0.8*19 = 19. e^{-4.\frac{22}{R} }$

$0.8 = e^{-\frac{88}{R} }$

$ln(0.8) = ln(e^{-\frac{88}{R} })$

$ln(0.8) = -\frac{88}{R}$

$R = -\frac{88}{ln(0.8)}$

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

7 0

3 years ago