4 years ago

What is the best way to avoid overloading your boat

Physics

1 answer:

8_murik_8 [283]4 years ago

6 0

Travel lightly and read your instructions manual.

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If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds

BabaBlast [244]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

According to above question ~

Current (I) = 4 Amperes

Time (t) = 3 seconds

Charge (q) = ?

Let's find the charge (q) by using formula ~

$I = \dfrac{q}{t}$

$4 = \dfrac{q}{3}$

$q = 4 \times 3$

$q = 12 \: \: coulombs$

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

7 0

3 years ago

An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k

alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

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3 years ago

Find the potential energy of a 2 kg ball 15 m in the air.

mart [117]

Answer:

294.3 Joules

Explanation:

2kg*9.81m/s^2*Δ15=294.3J

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3 years ago

Evaluate the scenarios and select the one that demonstrates the training principle overload.

matrenka [14]

Answer:

B :)

Explanation:

:) JUST TRUST ME I GOT IT CORRECT

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3 years ago

Read 2 more answers

A horse is trotting along pulling a sleigh through the snow. To move the 225 kg sleigh straight ahead at a constant speed, the h

RoseWind [281]

Answer:

a) The net force acting on the sleigh is zero because it is moving at a constant speed.

b) $F_{fr}= \mu mg = F_{pull}$ , then $\mu = \frac{F_{pull}}{mg} = 0.0557$ .

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3 years ago