All categories

2 years ago

Which of the objects below has the greatest acceleration? *

Physics

1 answer:

lesya692 [45]2 years ago

8 0

$\\ \bull\sf\dashrightarrow F=ma$

$\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}$

Now

$\\ \bull\sf\dashrightarrow a\propto\dfrac{1}{m}$

$\\ \bull\sf\dashrightarrow a\propto F$

Lower mass=Higher acceleration
Lower Force=Lower Acceleration

Option B has lowest mass and highest force hence its correct

You might be interested in

How does a vacuum flask keep drinks hot? Explain in terms of conduction, convection and radiation.

Flura [38]

Vacum flask has inner layer of glass and outer body of plastic.the glass has a reflective metal layer..when a hot drink is poured into the flask it prevents conduction.tight lid stops the convection.when radiation tries to leave the hot drink the reflective layer of the glass reflects it back so no heat can escape.

4 0

3 years ago

Is it possible for a circuit to be connected both in series and in parallel?

marshall27 [118]

Hello There!

Well, technically no. It has to be one or the other, it can't be both.

Hope This Helps You!
Good Luck :)

- Hannah ❤

4 0

3 years ago

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I E I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I E I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago

suppose you needed to build a raft to cross a fast-moving river. Describe the physical and chemical properties of the raft that

erastovalidia [21]

1.Density of the material building the raft must be lower than the water.
2. Material must not react with water. 
3.Material must have high strength. 
4.Raft must be wide in-order to avoid drawing in the river.

4 0

3 years ago

A 500 kg table with 4 legs rests on solid flat ground. I place 3 books on the center of the table with masses of 10 kg, 20 kg an

BARSIC [14]

Answer:

1,373.4 N

Explanation:

The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.

Mass of books will be 10kg+20kg+30kg=60 kg

Total mass of table and books will be 500kg+60kg=560 kg

This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg

Force is product of mass and acceleration due to gravity hence F=gm

Taking g as 9.81 m/s2 then

F=140*9.81=1,373.4 N

Therefore, rhe normal force is equivalent to 1,373.4 N

6 0

3 years ago