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lorasvet [3.4K]
3 years ago
10

The

Physics
1 answer:
KonstantinChe [14]3 years ago
8 0

Answer:

c. Velocity

Explanation:

Velocity is a measure of change of position of the body with time. Change of position means the distance traveled by the object in a given interval of time but in a particular direction.

We know that speed is given as the distance traveled over a given time interval without taking into account the direction of motion. So, the magnitude of velocity is same as its speed. Velocity, being a vector quantity, has both magnitude and direction.

Therefore, the <u>velocity</u> of an object consists of its speed and direction.

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The box is pushed to the right with a force of 40 Newtons and it just begins to move what is the maximum static frictional force
Vesnalui [34]

Answer:

The maximum static frictional force is 40N.

Explanation:

When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).

This coefficient defines the maximum static friction force that we can have.

So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.

In this case, we know that we apply a force of 40N and the object just starts to move.

Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.

3 0
3 years ago
Two resistors, R1 and R2, are connected
marusya05 [52]

Answer:

18 ohms

Explanation:

V = I(R1 + R2)

5V = (0.167A)(12 ohms + R2)

Solving for R2

R2 = 18 ohms

3 0
3 years ago
Which way does wind blow
prisoha [69]
I think it blows vertically and horizontally cause wind can blow different directions
5 0
3 years ago
Read 2 more answers
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0
3 years ago
Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra
vredina [299]

Answer:

At low pressure- \mu_{k}=0.02315

At high pressure- \mu_{k}=0.00445

Explanation:

Initial speed, V_{i}=3.3 m/s

Final speed, V_{f}=3.3/2= 1.65 m/s

Net horizontal force due to rolling friction F_{net}=\mu_{k} mg where m is mass, g is acceleration due to gravity, \mu_{k} is coefficient of rolling friction

From kinematic relation, V_{f}^{2}- V_{i}^{2}=2ad

For each tire,

V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd

Making \mu_{k} the subject

\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}

Under low pressure of 40 Psi, d=18 m

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315

Therefore, \mu_{k}=0.02315

At a pressure of 105 Psi, d=93.7

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445

Therefore, \mu_{k}=0.00445

4 0
3 years ago
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