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3 years ago

A car travels in a straight line for 4.4 h at a constant speed of 81 km/h. What is its acceleration?

Physics

2 answers:

Fed [463]3 years ago

7 0

The speed and length of time don't matter.

Constant speed in a straight line is the DEFINITION of ZERO acceleration.

irga5000 [103]3 years ago

5 0

Answer:

$a = 0m/s^2$

Explanation:

we can use the equation:

$v_f = a(t) + v_i$

And plug in our numbers:

$81km/h = am/s^2(4.4h) + 81hm/h$

$0 = am/s^2(4.4h)$

$0 = am/s^2$

$a = 0m/s^2$

Because the speed was constant, there was no difference between the final and initial velocities, meaning that once you do the math you see there is no acceleration affecting the speed.

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Your mom is making lunch which forces are acting on your mom

N76 [4]

The forces acting on your mom while cooking is Air resistance and the force of friction

Explanation:

1. Air resistance:

In simple words, Air resistance can be stated as the type of friction between the air and the other materials.
In this scenario, there will be an air resistance and the air hits the mom while cooking via the doors or windows

2. The force of friction:

In simple words, friction can be stated as, the resistance that one surface or object encounters when moving over another.
While cooking the food mom would experience the friction since friction is the transfer of heat, and cooking is the process of receiving that heat.

7 0

4 years ago

Read 2 more answers

What does the term mccarthyism describe? To whom does it refer?

kolbaska11 [484]

Mccarthyism describes a political witchhunt, specifically refers to Sen. McCarthy's cold war era congressional commitees whose aim was to investigate and "out" or reveal individuals with communist "sympathies".

6 0

4 years ago

What will be the value of both charges if they are 5 cm apart and suffer a

Mrac [35]

Answer:

$\boxed{q = 1.2 \times {10}^{ - 6} C}$

Explanation:

$f_e = \frac{{q}^{2}k }{ {r}^{2} } \\ q = \sqrt{ \frac{f_e( {r}^{2} )}{k} } = \sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } \\ q =\sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } = \sqrt{ \frac{0.013}{9 \times {10}^{9} } } \\ q = 1.2 \times {10}^{ - 6}$

3 0

3 years ago

When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static

natali 33 [55]

The coefficient of static friction is 0.234.

Answer:

Explanation:

Frictional force is equal to the product of coefficient of friction and normal force acting on any object.

So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.

Normal force = mass * acceleration due to gravity

Normal force = 2 * 9.8 = 19.6 N.

And the frictional force is given as 4.6 N, then

$Coefficient of static friction = Frictional force/Normal force$

Coefficient of static friction = 4.6 N / 19.6 N = 0.234

So the coefficient of static friction is 0.234.

3 0

3 years ago

Is there a link between number of bulbs and current drawn from the power pack?

Maurinko [17]

Answer:

Yes, there a link between number of bulbs and current drawn from the power pack.

Explanation:

In an Electrical circuit, we have resistors present in that circuit. These resistors can be connected in two ways.

a) Series connection

b) Parallel connection

There is a link or a relationship between number of bulbs and the current drawn from the power pack. This is because the number of bulbs is equivalent to or equal to the number of resistors.

Hence,

a) In a series connection, the link or relationship between the number of bulbs(resistors) is as the number of light bulbs increases, the current in the power pack (circuit) decreases.

b) In a parallel connection, the link or relationship between the number of bulbs(resistors) is as the number of light bulbs increases, the current in the power pack (circuit) increases.

5 0

3 years ago