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3 years ago

Which is the best estimate for the mass of a typical lineman on a high school football team

Physics

1 answer:

fgiga [73]3 years ago

7 0

We have that the best estimate for the mass of a typical lineman on a high school football team is

$100,000 g=100kg$

From the options

1. 100,000 g

2. 100,000 kg

3. 100,000 mg

Generally

A Typical lineman cannot be as small as

$100000mg=0.1kg=100g$

A Typical linesman cannot be as massive as

$100,000kg$ as that is unrealistic

With the Average Mass of A Typical linesman can be

$100,000 g=100kg$

In conclusion

The best estimate for the mass of a typical lineman on a high school football team is

$100,000 g=100kg$

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A can of tuna, starting from rest, is dropped off a building. If it is only being pulled down by gravity (-9.8 m/s2) and it hits

tia_tia [17]

Answer:

Explanation:

The distance talks about how far an object has travelled.

Given

9 = -9.8m/s²

t = 3.2s

to get the distance Δx, we will use the formula;

S = ut+1/2gt²

S = 0(3.2)+1/2(-9.8)(3.2)²

S = 0-4.905(10.24)

S = -50.23m

Hence the can of tuna drop by 50.23m

3 0

3 years ago

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a hor

Leno4ka [110]

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

7 0

3 years ago

The driver of a car applies the brakes for 4.0 s.As a result an average force of 7500N is exerted in the car to slow it down.Wha

Sindrei [870]

Answer:

The change in momentum of the car is 30,000 kg m/s

Explanation:

Given,

The force exerted on the car to slow down, F = 7500 N

The time period of force, t = 4 s

The rate of change of momentum of the object is equal to the force acting on it.

Therefore,

(mv - mu) / t = F

Where v and u are the final and initial velocity of the car. The change in momentum of the car,

mv - mu = F x t

= 7500 x 4

= 30,000 kg m/s

Hence the change in momentum of the car is 30,000 kg m/s

7 0

3 years ago

A circuit contains two light bulbs connected in parallel. What would happen to the brightness of each light bulb if two more lig

DENIUS [597]

Answer:

The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.

Explanation:

Brightness of the bulb is given as

$P= \frac{V^2}{R}$

since all bulbs are connected in parallel so here voltage across each bulb will remain same and resistance of each bulb is "R"

So here power across each bulb will remain the same always.

So there will be no effect on the power or brightness of bulb.

Now we also know that equivalent resistance is given as

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}............$

$R_{eq} = \frac{R}{n}$

so here equivalent resistance will decrease on adding more resistance in parallel.

so correct answer will be

The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.

6 0

3 years ago

Read 2 more answers

If the pipe of the previous exercise is connected to a hose and the speed there is 3.8 m/s, what is the diameter of the hose?

Amiraneli [1.4K]

The diameter of the hose is 6.34 cm.

"Your question is not complete, it seems to be missing the following information";

the flow rate of water in the pipe is 0.012 m³/s

The given parameters;

velocity of water in the hose, v = 3.8 m/s

flow rate of water in the hose, Q = 0.012 m³/s

Volumetric flow rate is directly proportional to the product of the area of the hose through which the water flows and the velocity of the water flowing through the hose.

Q = Av

where;

Q is the volumetric flow rate

A is the area of the hose

v is the velocity of flow

The area of the hose is calculated as follow;

$A = \frac{Q}{v} \\\\A = \frac{0.012}{3.8} \\\\A = 0.00316 \ m^2$

The diameter of the hose is calculated as follows;

$A = \frac{\pi D^2}{4} \\\\\pi D^2 = 4A\\\\D^2 = \frac{4 \times A}{\pi} \\\\D = \sqrt{\frac{4 \times A}{\pi} } \\\\D = \sqrt{\frac{4 \times 0.00316}{\pi} } \\\\D = 0.0634 \ m\\\\D = 6.34 \ cm$

Thus, the diameter of the hose is 6.34 cm.

Learn more here: brainly.com/question/15061170

4 0

2 years ago