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Marianna [84]
3 years ago
11

What type of current is produced by a battery?

Physics
2 answers:
nadya68 [22]3 years ago
8 0
The type of current produced by a battery is direct current. C.
melamori03 [73]3 years ago
8 0
The answer is C direct current
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Compared to a 1-kg block of solid iron, a 2-kg block of solid iron has the same:
algol13

The Correct choice is :

  • Density
6 0
3 years ago
a truck of mass 2 500 kg travelling at the speed of 20 ms–1 to the right collide head-on with another car of mass 1 000 kg. If b
monitta

Answer:

50 m/s opposite direction to the motion of the truck

Explanation:

From the question,

Applying the law of conservation of momentum

mu+m'u' = V(m+m')...….. Equation 1

Where m = mass of the truck, u = initial velocity of the truck, m' = mass of the car, u' = initial velocity of the car, V = Final velocity after collision

Given: m = 2500 kg, u = 20 m/s, m' = 1000 kg, V = 0 m/s (both car stop after collision)

Substitute these values into equation 1

2500(20)+1000(u') = 0(2500+1000)

2500(20)+1000(u') = 0

Solve for u'

u' = -[2500(20)]/1000

u' = -50 m/s

The negative sign shows that the car travels in opposite direction to the truck

Hence the car initial velocity before collision is 50 m/s in opposite direction to the motion of the truck

4 0
2 years ago
Is it possible for a baseball to have as large a momentum as a much more massive bowling ball
kirza4 [7]
Yes it is possible. Momentum is calculated by the mass of the object times its velocity.
For example, say a bowling ball weighs 3.0kg and is travelling at a speed of 3.0m/s. Its momentum would be 3.0×3.0=9.0 kg·m/s.
Now say we have a baseball weighing 0.20kg and it is travelling at a speed of 47.0m/s. Its momentum would be 0.20×47.0=9.4 kg·m/s, which is more than that of the bowling ball.
8 0
3 years ago
A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.10 s. The particle has a charge
mamaluj [8]

Answer:

The magnitude of the electric field is 8.6\times10^{2}\ N/C

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

a=\dfrac{v-u}{t}

Put the value into the formula

a=\dfrac{160-0}{2.10}

a=76.19\ m/s^2

We need to calculate the magnitude of the electric field

Using formula of electric field

E=\dfrac{F}{q}

E=\dfrac{ma}{q}

E=\dfrac{a+g}{\dfrac{q}{m}}

Put the value into the formula

E=\dfrac{76.19+9.8}{0.100}

E=8.6\times10^{2}\ N/C

The direction is upward.

Hence, The magnitude of the electric field is 8.6\times10^{2}\ N/C

4 0
3 years ago
Read 2 more answers
A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt
anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

6 0
3 years ago
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