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strojnjashka [21]
3 years ago
6

Please help on this one?

Physics
1 answer:
olganol [36]3 years ago
6 0

the object distance of both lenses are positive.

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An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the
alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

5 0
3 years ago
An electric toy with a resistance of 2.50 Ω is operated by a 3.00-V battery. (a) What current does the toy draw? (b) Assuming th
Leno4ka [110]

Answer:

a) The current is i = 1.2 A

b) The charge is Q = 17280 C

c) The energy is E = 43200 J

Explanation:

a) The current is given by the ohm's law wich is:

i = V/R = 3/2.5 = 1.2 A

b) Since the charge is steady we can use the following equation to find the charge amount in that time:

i = Q/t

Q = t*i

Where t is in seconds, so we have 4h * 3600 = 14400 s

Q = 1.2*14400 = 17280 C

c)  The energy is the power delivered to the toy multiplied by the time:

P = 1.2*2.5 = 3 W

E = P*t = 3*14400 = 43200 J

7 0
3 years ago
WILL GIVE BRAINLIEST AND 30 POINTS!
mariarad [96]

Answer:

Gallium

Explanation:

Gallium is one such element used as a do/pant in a p-type semiconductor.

A do/pant is an impurity added to a semi-conductor used to alter its properties. Semi-conductors have a wide range of applications. They will conduct heat and electricity only under certain conditions. This property is highly desirable and find a wide application in electronics.

For p-type conductors, they are best do/ped with elements with 3 valence electrons. These are group 3 elements. From the choices, only gallium belongs to this group.

Other elements given are good do/pants for n-type semiconductors. They have 5 valence electrons.

8 0
3 years ago
calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s​
zheka24 [161]

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

5 0
2 years ago
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi
natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0
3 years ago
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