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3 years ago

How does increasing the tension of a spring affect a wave on the spring?

2 answers:

4 0

Increasing the tension of a spring affects a wave on the spring because it increases the frequency. When the tension rises, so does the frequency.

mezya [45]3 years ago

3 0

Answer: Increases the frequency of the wave

Explanation:

Increasing the tension of a spring means increasing its stiffness. The wave produced thereafter has greater frequency.

Higher tension causes the spring to have greater restoring force causing it to quickly come back to its equilibrium position. Thus, the speed of the wave also increases.

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The temperature of a gas is increased from 125 celsius inside a rigid container. The original pressure of a gas was 1.22atm, wha

Sonja [21]

The temperature of a gas is increased from 125 celsius inside a rigid container. The original pressure of a gas was 1.22atm, what will the pressure of a gas be after the temperature changes?

8 0

3 years ago

Find the potential energy of a 40kg cart that is on a hill that is 10 meters high

Vilka [71]

Explanation:

P.E. = mgh

PE = 40 × 10 × 10

PE = 4000 Joule

3 0

2 years ago

One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what

Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

$F_y=ma_y$ (2)

$F_x=ma_x$ (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

$F_y=W+n=0$

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

$F_y=(755)(-9.81)+n=0$

$n=7406.55 N$ (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

$F_x=F+f=ma_x$

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

$1988+f=m(1.36)$

$f=(755)(1.36)-1988=-961.2 N$

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

$|f|=\mu_k n$

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

$\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13$

4 0

3 years ago

What is the car's acceleration from 0 to 1 second?

dusya [7]

10 mph/s because there is 60 seconds in a minute then divide by 6 which is 10.

3 0

3 years ago

Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10

nydimaria [60]

Answer:

Explanation:

The formula for this is

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

$7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2}$ and moving things around to solve for r:

$r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} }$ Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

8 0

3 years ago