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Leni [432]
2 years ago
15

If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be?

Physics
1 answer:
kicyunya [14]2 years ago
6 0

For this case we have that by definition:

F = ma

Where,

  • <em>m: mass of the object </em>
  • <em>a: acceleration of the object </em>
  • <em>F: summation of forces </em>

We have then:

F = 10 + 5\\F = 15 N

Then, by clearing the acceleration we have:

a = \frac {F} {m}

Substituting values we have:

a = \frac {15} {5}\\a = 3 \frac {m} {s ^ 2}

Answer:

The acceleration of the box is equal to:

a = 3 \frac {m} {s ^ 2}

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If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
Anna35 [415]

This question is incomplete

Complete Question

Three equal point charges are held in place as shown in the figure below

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

A) F1=2F2

B) F1=3F2

C) F1=4F2

D) F1=9F2

Answer:

D) F1=9F2

Explanation:

We are told in the question that there are three equal point charges.

q, Q1, Q2 ,

q = Q1 = Q2

From the diagram we see the distance between the points d

q to Q1 = d

Q1 to Q2 = nd

Assuming n = 2

= 2 × d = 2d

Sum of the two distances = d + 2d = 3d

F1 is the force on q due to Q1 and

F2 is the force on q due to Q2,

Since we have 3 equal point charges and a total sum of distance which is 3d

Hence,

F1 = 9F2

6 0
2 years ago
If the frequencies of two component waves are 24 Hz and 20 Hz, they should produce _______ beats per second.
horrorfan [7]
This can be answered using the beat frequency formula, which is simply the difference between 2 frequencies.

Let: <span>fᵇ = beat frequency
</span>f₁ = first frequency
f₂ = second frequency

fᵇ = |f₁ - f₂|

substituting the values:
fᵇ = |24Hz - 20Hz|
fᵇ = 4Hz

The unit Hz also means beats per second, therefore:
<span>fᵇ = 4 beats per second
</span>
Therefore, the answer is C. 4
8 0
2 years ago
Read 2 more answers
A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a
vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451

tan\alpha = 0.451

\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0

3 0
2 years ago
Which pair of sentences is describing the same velocity? A car is parked. A car is moving in circles. A bus drives 40 miles per
olya-2409 [2.1K]
The bus and the truck have the same velocity.

Also, Valerie and Owen have the same velocity.
8 0
2 years ago
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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
1 year ago
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