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2 years ago

What fraction of the horse's weight is borne by the front hooves?

Physics

1 answer:

Gala2k [10]2 years ago

3 0

Answer: 3056N

Explanation:

Torque =rF

Since left side of the torque=right side of the torque

1.1(5000-x)=(1.8-1.1)(x)

x(weight of the front side), x=3055.56

Fraction of the horses weight = (weight of the front side)/(total weight)

1.8m/1.1m = 1.636m

Then, 5000N/1.636m =3056N

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An electric drill transfers 200 J of energy into a useful kinetic energy store. It also transfers 44 J of energy by sound and 48

kramer

Answer:

Er = 108 [J]

Explanation:

To solve this problem we must understand that the total energy is 200 [J]. Of this energy 44 [J] are lost in sound and 48 [J] are lost in heat. In such a way that these energy values must be subtracted from the total of the kinetic energy.

200 - 44 - 48 = Er

Where:

Er = remaining energy [J]

Er = 108 [J]

3 0

2 years ago

The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the

satela [25.4K]

The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
$I= \frac{P}{A}$ (1)
where
P is the power
A is the area

In our problem, the intensity is $I=6.0 \cdot 10^{-10} W/m^2$ . At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
$A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2$

And so if we re-arrange (1) we find the power emitted by the source:
$P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W$

3 0

2 years ago

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete

Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(15)^2}{22.5}$

$a=10\ m/s^2$

We know that, $g=9.8\ m/s^2$

$a=\dfrac{10\times g}{9.8}$

$a=1.02\ g$

Hence, this is the required solution.

5 0

2 years ago

Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another

ioda

Answer:

$6.1875\times 10^{-8}$

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies $I=\frac {k}{r^{2}}$ where r is the distance

Since intensity is given then when at 3 m

$1.1\times 10^{-7}= \frac {k}{3^{2}}$

$k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}$

Since we have the constant then at 4m

Intensity, $I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}$

8 0

2 years ago

One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra

sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration ( $a_{r}$ ), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

$a_{r} = 4\pi^{2}\cdot f^{2}\cdot R$ (1)

Where:

$f$ - Frequency, measured in hertz.

$R$ - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension ( $T$ ), measured in newtons, through the string, then we derive the following model:

$\Sigma F = T = m\cdot a_{r}$ (2)

Where $m$ is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

$T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R$ (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

$\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right)$ (4)

$\frac{T_{2}}{T_{1}} = 4\cdot 2$

$\frac{T_{2}}{T_{1}} = 8$

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0

2 years ago