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alina1380 [7]
3 years ago
8

Consider the relationship of the variables in Newton’s second law. In a drag car race, the force applied to the car is doubled b

y the driver stepping on the gas pedal. The acceleration of the car will . The mass of the car will . The velocity of the car will .
Physics
2 answers:
kkurt [141]3 years ago
5 0

When the driver mashes the gas, puts the pettle to the mettle, and doubles the engine force applied to the car . . .

-- the acceleration of the car doubles

-- the mass of the car doesn't change

-- the velocity of the car starts to increase in the direction in which the car is already moving.  If the car's speed was already increasing forward before the force adjustment, then the rate at which it increases due to the doubled force will double.

daser333 [38]3 years ago
4 0

Recall that the force on an object is related to the mass and acceleration of that object by the formula F = ma, where m is the mass of the object and a is its acceleration. What happens when we double F? Well, you might remember from algebra that, in order to keep our equality true, if we double one side, we must also double the other, so our equation becomes 2F = 2ma. Now, this means one of two things: either the mass has doubled, or the acceleration has doubled.

We can tell right away that it'd be absurd if a race car doubled in mass every time it hit the gas, so the quantity doubling must be the <em>acceleration. </em>If we call the car's current velocity v1, we'll be adding the doubled acceleration to get its new velocity. Mathematically, v = v1 + 2a.

We can now conclude that, by doubling the force:

  • The acceleration of the car will double,
  • The mass of the car will stay the same, and
  • The velocity of the car will increase by double the original acceleration
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Two observers are 300 ft apart on opposite sides of a flagpole. The angles of elevation from the observers to the top of the pol
valentina_108 [34]

Answer:

h = 48.077 ft

Explanation:

given,

distance between two observer = 300 ft

angle of elevation to top pole = 16° and 20°

height of the flagpole = ?

now,

Let h be the height of the flagpole

Let x be the distance of the pole

tan 16^0 = \dfrac{h}{x}

x =\dfrac{h}{tan 16^0}

now,

again applying

tan 20^0 = \dfrac{h}{300-x}

300-x=\dfrac{h}{tan 20^0}

300-3.49 h=2.75 h

6.24h = 300

h = 48.077 ft

3 0
3 years ago
Resistors and reactors, for use over 600 volts, shall not be installed in close enough proximity to combustible materials to con
suter [353]

Resistors and reactors, for use over 600 volts, shall not be installed in close enough proximity to combustible materials to constitute a fire hazard and shall have a clearance of not less than<u> 300 mm </u>from combustible materials.

Explanation:

  • The hazards associated with high power industrial resistors are primarily due to their open construction, which is necessary for cooling.
  • The exposed conductors which make up the resistors can be not only a shock hazard but also a thermal burn hazard.
  • When a resistor fails, it either goes open or the resistance increases. When the resistance increases, it can burn the board, or burn itself up.
  • Avoid touching non-flammable resistors in operation; the surface temperature ranges from approximately 350 °C to 400°C when utilized at the full rated value. Maintaining a surface temperature of 200°C or less will extend resistors service life.
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3 years ago
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F=g(m1*m2)/r^2
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7 0
3 years ago
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A 5.0 cm object is 6.0 cm from a convex lens, which has a focal length of 7.0 cm.
Reil [10]

Answer : The distance of the image from the lens is, -42 cm

The height of the image is, 35 cm

Solution :

First we have to calculate the image distance.

Formula used :

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where,

f = focal length = 7 cm

p = object distance = 6 cm

q = image distance = ?

Now put all the given values in the above formula, we get the distance of the image from the lens.

\frac{1}{7cm}=\frac{1}{6cm}+\frac{1}{q}

q=-42cm

Therefore, the distance of the image from the lens is, 42 cm and the negative sign indicates that the image is virtual.

Now we have to calculate the height of the image.

Formula used :

\frac{h}{h'}=\frac{p}{q}

where,

h = height of object = 5 cm

h' = height of image = ?

Now put all the given values in this formula, we get the height of the image.

\frac{5cm}{h'}=\frac{6cm}{-42cm}=-35cm

Therefore, the height of the image is, 35 cm and the negative sign indicates that the image is inverted.

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4 years ago
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dmitriy555 [2]
Answer = C

It blows towards the poles from the sub-tropical high pressure areas
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