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2 years ago

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3. When the procedure is repeated with a third line how will it distinguish whether the location of the center of gravity is acc

urate or not?

1 answer:

Law Incorporation [45]2 years ago

4 0

Well this is question is easy. I mean i’m the one to say the least, The answer is.. SIKE L bozo imagine not knowing the answer

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A horizontal force acts on an object on a fric- tionless horizontal surface. If the force is halved and the mass of the object i

guapka [62]

(a) If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b) if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

$\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}$

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

$\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)$

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

$\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)$

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

Learn more here:brainly.com/question/19887955

3 0

2 years ago

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm

lina2011 [118]

Answer:

Difference in height = 7.5 cm

Explanation:

We are given;.

Height of ethyl alcohol;h2 = 20 cm = 0.2 m

Density of glycerin: ρ1 = 1260 kg/m³

Density of ethyl alcohol; ρ2 = 790 kg/m³

To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.

Thus;

P1 = P2

Formula for pressure is; P = ρgh

Thus;

ρ1 × g × h1 = ρ2 × g × h2

g will cancel out to give;

ρ1 × h1 = ρ2× h2

Making h1 the subject, we have;

h1 = (ρ2× h2)/ρ1

h1 = (790 × 0.2)/1260

h1 = 0.125 m

Difference in height will be;

Δh = h2 - h1

Δh = 0.2 - 0.125

Δh = 0.075 m = 7.5 cm

4 0

3 years ago

How fast is Barry Allen ( The Flash) in season 5 of the Flash?

Burka [1]

Answer: The Flash, Allen's top speed is Mach 3.3, or 2,532 miles per hour.

Explanation:

8 0

3 years ago

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

3 0

3 years ago

Read 2 more answers

Q1: A cyclist brakes to a stop. His thinking distance was 1m and his braking distance was 3m. What was his overall stopping dist

weeeeeb [17]

Answer:

1.) 4m

2.) 37 m

3.) 62m

4.) 2.5 s

Explanation:

1.) Given that the

Thinking distance = 1m

Breaking distance = 3m

Stopping distance = breaking distance + thinking distance

Stopping distance = 1 + 3 = 4m

2.) Given that the

Stopping distance = 52 m

Thinking distance = 15m

Breaking distance = 52 - 15 = 37m

3.) The stopping distance = 76m

Thinking distance = 14m

Breaking distance = 76 - 14 = 62m

It take the brakes 62m to slow the car down to a stop.

4.) Given that a lorry travels 28m when stopping from a speed of 4m/s. If its braking distance was 18m, what was the driver’s reaction time?

Thinking = stopping distance - braking distance

Thinking distance = 28 - 18 = 10m

Speed = distance/time

4 = 10/reaction time

Reaction time = 10/4

Reaction time = 2.5 s

5.) Question incomplete

5 0

3 years ago