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2 years ago

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Which of the following represents a possible magnitude for the force of static friction when Xavier applied 72.1 Newtons of forc

e on the cart?

1 answer:

lana66690 [7]2 years ago

8 0

The possible magnitude for the force of static friction on the stationary cart is 72.1 N.

The given parameters:

<em>Applied force on the cart, F = 72.1 N</em>

<em />

Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.

F = ma

Static frictional force is the force resisting the motion of an object at rest.

$\Sigma F = 0\\\\F -F_f = 0$

where;

$F_f$ is the frictional force

$F= F_f \\\\72.1 = F_f\\\\F_f = 72.1\ N$

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.

Learn more about Newton's second law of motion: brainly.com/question/25307325

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A car initially traveling at 15.0 m/s accelerates at a constant rate of 4.50 m/s2 over a distance of 45.0 m. How long does it ta

anygoal [31]

Answer:

2.24 seconds

Explanation:

xf = xo + vo t + 1/2 at^2

45 = 0 + 15 t + 1/2 (4.5) t^2

2.25 t^2 + 15t - 45 = 0 Quadratic formula shows t = 2.24 seconds

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2 years ago

A physics student tests the theory of projectile motion by leaping off a 225 meter tall building. She runs off the building hori

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In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,

d = V₀t + 0.5gt²

Substituting the known values,

225 = (0 m/s)(t) + (0.5)(9.8)(t²)

Simplifying,

t = 6.776 s

To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.

range = (12.5 m/s)(6.776 s)

range = 84.7 m

<em>Answer: 84.7 m</em>

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3 years ago

Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is

s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: $\frac{∆v}{t}$

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 $\frac{feet}{second}$ then sped up to 5 $\frac{feet}{second}$ , then the change in my speed would be 2 $\frac{feet}{second}$ , because I started walking 2 $\frac{feet}{second}$ faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 $\frac{meters}{second}$ to 16 $\frac{meters}{second}$ in 4 seconds. What would the acceleration be? Let's set up an equation:

a = $\frac{∆v}{t}$

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 $\frac{meters}{second}$ by 4 $\frac{meters}{second}$ . That makes sense, right? Back to the equation.

a = $\frac{∆v}{t}$
a = $\frac{16-4}{4}$

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = $\frac{12}{4}$

a = 3 ( $\frac{meters}{second^{2}}$ )

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit $\frac{\frac{meters}{second}}{second}$ , which can be simplified down to $\frac{meters}{second^{2} }$ )

Let me know if you need clarification on anything I explained here!
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2 years ago

Your cell phone typically consumes about 390 mW of power when you text a friend. If the phone is operated using a lithium-ion ba

yulyashka [42]

Answer:

I = 0.11 A

Explanation:

In an electric circuit, the power delivered to a load, is just the product of the potential difference between the load terminals, times the current flowing through it, as follows:

$P = V*I$

In this case, the power is the one consumed by the cell phone = 390 mW, and the voltage the one produced by the internal energy of the battery, 3.5 V, neglecting the voltage loss at the internal resistance of the battery.
So, we can solve the above equation for the current I, as follows:

$I = \frac{P}{V} = \frac{0.39W}{3.5V} = 0.11 A$

The current flowing through the cell-phone circuitry is 0.11 A.

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3 years ago

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

Explanation:

(a) Expression for change in temperature is as follows.

$|\Delta T| = |x - y|K$

= 15.1 K

= $|(x - 273.15) - (y - 273.15)|^{o}C$

= $|x - y|^{o}C$

= $15.1^{o}C$

Therefore, the magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) Change in temperature from Celsius to Fahrenheit is as follows.

F = 1.8C + 32

C = $\frac{F - 32}{1.8}$

Since, K = C + 273

or, $\Delta K = \Delta C = \frac{\Delta F}{1.8}$

$\Delta F = 1.8 \Delta K$

= 1.8 (15.1)

= $27.18^{o}F$

or, = $27.2^{o}F$

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

7 0

3 years ago

Read 2 more answers