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9966 [12]
2 years ago
8

Suppose you had an electrical circuit and decided to change it into a new circuit. does the equivalent resistance of the new cir

cuit compare to the old one ? If you were to insert additional resistors in series, how 。
a. No effect on the equivalent resistance
b. The equivalent resistance decreases
c. The equivalent resistance increases
Physics
1 answer:
Fittoniya [83]2 years ago
7 0

Answer:

Yes becuase the circuit is fryed like sum french fries so yes yes is the yes option yes okay, yes but no

Explanation:

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A sled is pushed to right with a force of 200 N. The force of friction opposing that motion is 50 N. a) What is the Net Force on
marin [14]

Answer:

Fr = 150 [N]

a = 5 [m/s²]

Explanation:

In order to find the resulting force, we must assume that the thrust force is positive to the right, while the friction force is negative to the left.

F_{r}=200-50\\F_{r}=150[N]

Now Newton's Second Law tells us that the sum of the forces or the resulting force is equal to the product of mass by acceleration.

F = m*a

150 = 30*a\\a=150/30\\a = 5 [m/s^{2} ]

4 0
3 years ago
A space rover weighs less on Mars than it does on Earth. Which statement explains this difference? A. The gravitational constant
Maksim231197 [3]

Answer:

B. The mass of Mars is less than the mass of Earth.

Explanation:

Mass of an object is the constant anywhere in the universe.

The weight of an object is equal to the gravitational force acting on it.

Weight is given by

W=\dfrac{GMm}{R^2}\\\Rightarrow W=\dfrac{GM}{R^2}m\\\Rightarrow W=mg

where

G = Gravitational constant

M = Mass of Planet

R = Radius of planet

m = Mass of object

g = Acceleration due to gravity

So weight of an object depends on the acceleration due to gravity on that planet. The acceleration due to gravity depends on the mass and radius of the planet.

The weight of the object is less on Mars because mars has less mass compared to Earth.

6 0
3 years ago
If it requires 7.0 J of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be re
belka [17]

Answer:

56 J

Explanation:

The following data were obtained from the question:

Energy 1 (E₁) = 7 J

Extention 1 (e₁) = 1.8 cm

Extention 2 (e₂) = 1.8 + 3.6 = 5.4 cm

Energy 2 (E₂) =?

Energy stored in a spring is given by the following equation:

E = ke²

Where E is the energy.

K is the spring constant.

e is the extension.

E = ke²

Divide both side by e²

K = E/e²

Thus,

E₁/e₁² = E₂/e₂²

7/ 1.8² = E₂/ 5.4²

7 / 3.24 = E₂/ 29.16

Cross multiply

3.24 × E₂ = 7 × 29.16

3.24 × E₂ = 204.12

Divide both side by 3.24

E₂ = 204.12 / 3.24

E₂ = 63 J

Thus, the additional energy required can be obtained as follow:

Energy 1 (E₁) = 7 J

Energy 2 (E₂) = 63 J

Additional energy = 63 – 7

Additional energy = 56 J

8 0
3 years ago
A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnit
dimaraw [331]

Answer:

e) 67°

Explanation:

the force on the wire can be calculated using the expression below

F = BILsinФ

But we are looking for the angle between the wire segment and the magnetic field, then we can make Ф the subject of the formula from the above expresion, then we have,

Ф =sin⁻¹ (F/BIL)

The parameters is defined as

I =current that is been carried by the wire= 0.35 A

Ф = angle between the wire segment and the magnetic field, which is the unknown?

L = length of the wire=65 cm

B = magnetic field = 1.24

F= force on the wire = 0.26 N,

Ф =sin⁻¹ (F/BIL)

Ф =sin⁻¹ X .....................eqn(#)

Where X= (F/BIL)

We can calculate for X= (F/BIL), from eqn(#) by substituting value of Force, Lenght and

magnetic field

X=(F/BIL)= 0.26/(1.24×0.35×0.65)

= 0.26/0.2821

=0.922

Then substitute X into eqn (Ф =sin⁻¹ X)

Then

Ф =sin⁻¹ (0.922)

Ф=66.42°

Ф=67° approximately

Therefore, the angle between the wire segment and the magnetic field is 67°

7 0
3 years ago
Can you name three dangers to the environment that some people use everyday in their homes without realizing it?
konstantin123 [22]

Answer:

1 COSMETICS WITH MICROBEADS

2 COFFEE CAPSULES

3 WET WIPES

Explanation:

1.Plastic microbeads began to be introduced in cosmetics, toiletries and cleaning products in the 1970s, but it wasn’t until the 1990s that brands began to incorporate them in a massive way. Toothpastes, creams, lotions, shampoos and detergents began to include microbeads as the great innovation of the time for achieving an abrasive effect, replacing the nature-based materials in use until then.

2 Capsule coffee machines have revolutionized the breakfasts of millions of people, offering a convenient and practical option for preparing an espresso or latte with the same quality one finds in a café. Data from 2017 indicate that 29% of US coffee consumers use these machines, a figure that continues to grow.

3They started off being an invaluable aid for fathers and mothers when facing the messy moment of their baby’s diaper change, but soon they began to be reinvented as deodorants, cleansers, disinfectants and hand soap substitutes, and even as toilet paper for adults. Wet wipes have become a common item in many homes, but with a dramatic consequence: they help to create fatbergs, immense accumulations that block the sewerage networks and that are composed of 93% non-degradable wipes, together with fat, condoms and other similar items thrown into the toilet.

3 0
3 years ago
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