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3 years ago

Which dog has the most kinetic energy?

Physics

2 answers:

Nadya [2.5K]3 years ago

5 0

A. A dog of mass 14kg running with the speed of 3m/s

Furkat [3]3 years ago

3 0

Nobody actually needs to know how much kinetic energy any of these dogs has.

The purpose of this question is to help you discover whether you know how to calculate kinetic energy, and then to give you some practice at doing it.

You have to calculate the kinetic energy of each dog, and then review your results to see which dog has the most.

The whole secret behind working this problem is the formula for the kinetic energy of any object:

Kinetic Energy = (1/2) · (mass) · (speed)²

The little ² next to 'speed' means 'squared' or 'multiplied by itself'.

(speed)² means (speed · speed) .

So, in order to calculate how much kinetic energy an object has, here's what you do:

-- Take the object's speed. Multiply the speed by itself.

-- Take that number and multiply it by the object's mass.

-- Take that number and throw away half of it.

So now here we go:

Dog A: KE = (1/2) (14 kg) (2 m/s)² = 28 Joules

Dog B: KE = (1/2) (14 kg) (3 m/s)² = 63 Joules

Dog C: KE = (1/2) (11 kg) (5 m/s)² = 137.5 Joules

Dog D: KE = (1/2) 12 kg) (4 m/s)² = 96 Joules

==> Dog C has more kinetic energy than any of the other 3 dogs.

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Shari uses the data in the table to predict how her weight on Venus would compare with her weight on Earth.

spin [16.1K]

Answer : Her weight is lower on Venus because the acceleration due to gravity is lower.

Explanation :

Venus is also called as Earth's twin. This is because both the mass and the size of Earth and Venus are almost same. The acceleration due to gravity on earth is $9.8\ m/s^2$ while on Venus is $8.87\ m/s^2$ .

So, when Shari measure her weight on Venus she found her weight is lower on Venus. This is because the acceleration due to gravity is lower on the surface of Venus as compared to the Earth.

Since, $w=mg$

i.e. weight depends on g.

So, correct prediction is (b)

3 0

3 years ago

Read 2 more answers

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)

xo = 10 km

v1 = 55km/h

car 2:

$x'=v_2t$ (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

$x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}$

$t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s$

The position in which both cars coincides is:

$x=(55km/h)(0.33h)=18.33km$

6 0

3 years ago

Which method of separation would work on a homogeneous mixture salt water

Butoxors [25]

To separate a mixture of salt and water, you can try first by using filter paper hen with the extra water part set it out to the window so that the salt water evaporates and only the salt is remaining.

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3 years ago

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Madelyn wanted to learn what sorts of things around the house are magnetic. She took a magnet around and touched it to many thin

madreJ [45]

Answer:

Explanation:

i just finished a chapter about this

7 0

3 years ago

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

$\boxed{B = 0 T}$

3 0

2 years ago