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3 years ago

Q3.

Physics

2 answers:

Anna35 [415]3 years ago

6 0

Answer:

A. Linear

Explanation:

hope I help you

lakkis [162]3 years ago

3 0

Answer:

Linear

Explanation:

The slope is constant.

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Say you have a differential drive robot that has an axle length of 30cm and wheel diameter of 10cm. Find the angular velocity fo

Klio2033 [76]

Answer:

a) ω1 = 18rpm ω2 = -18rpm

b) ω1 = 102rpm ω2 = 138rpm

c) ω1 = ω2 = 3.18rpm

Explanation:

For the first case, we know that each wheel will spin in a different direction but with the same magnitude, so:

ωr = 6rpm This is the angular velocity of the robot

$\omega = \frac{\omega r * D/2}{r_{wheel}}$ where D is 30cm and rwheel is 5cm

$\omega = \frac{6 * 30/2}{5}=18rpm$ One velocity will be positive and the other will be negative:

ω1 = 18rpm ω2 = -18rpm

For part b, the formula is the same but distances change. Rcircle=100cm:

$\omega 1 = \frac{\omega r * (R_{circle} - D/2)}{r_{wheel}}$

$\omega 2 = \frac{\omega r * (R_{circle} + D/2)}{r_{wheel}}$

Replacing values, we get:

$\omega 1 = \frac{6 * (100 - 30/2)}{5}=102rpm$

$\omega 2 = \frac{\omega r * (100 + 30/2)}{5}=138rpm$

For part c, both wheels must have the same velocity:

$\omega = \frac{V_{robot}}{r_{wheel}}=20rad/min$

$\omega = 20rad/min * \frac{1rev}{2*\pi rad}=3.18rpm$

8 0

4 years ago

What energy change takes place when the wrecking ball strikes the wall?

Ilia_Sergeevich [38]

Answer:

Explanation:

The wrecking ball transfers kinetic energy to the wall.

3 0

2 years ago

A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What m

Goryan [66]

Answer:

ΔE = GMm/24R

Explanation:

centripetal acceleration a = V^2 / R = 2T/mr

T= kinetic energy

m= mass of satellite, r= radius of earth

= gravitational acceleration = GM / r^2

Now, solving for the kinetic energy:

T = GMm / 2r = -1/2 U,

where U is the potential energy

So the total energy is:

E = T+U = -GMm / 2r

Now we want to find the energy difference as r goes from one orbital radius to another:

ΔE = GMm/2 (1/R_1 - 1/R_2)

So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R

ΔE = GMm/2R (1/3 - 1/4)

ΔE = GMm/24R

6 0

3 years ago

Which statement correctly compares weight and mass?

Varvara68 [4.7K]

Answer:

Option D. Weight varies with location, but mass does not.

Explanation:

To know which option is correct, it is important that we have a background knowledge of mass and weight.

A brief summary of the difference between mass and weight is given below:

1. Mass is the quantity of matter present in an object while weight is the gravitational pull on an object.

2. The SI unit of mass is kilogram Kg) while that of weight is Newton (N)

3. Mass is constant while weight varies by location.

4. Mass can measured using a chemical balance or beam balance while weight can be measured using a lever or spring balance.

With the above information, we can see that mass of an object is always but the weight varies by location.

6 0

4 years ago

Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that

Greeley [361]

Answer:

The value is $m_i = 0.0234 \ kg$

Explanation:

Generally from the calorimetry principle we have that

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

So here heat gained water is mathematically represented as i.e

$Q_w = m_w * c_w * (T_w - T )$

substituting 0.150 kg for $m_w$ , 4200 J/kg.°C for $c_w$ , 100°C for $T_w$ and 75°C for T

We have

$Q_w = 0.150 * 4200 * (100 - 75 )$

$Q_w =15750 \ J$

The Heat loss by the ice is mathematically represented as

$Q_i = Q_1 + Q_2 + Q_3$

Here $Q_1$ is the energy to move the ice to its melting point which is evaluated as

$Q_1 = m_i * c_i * ( T_o -T_i)$

Here $m_i$ is the mass of ice

$c_i$ is the specific heat of ice with value $2.05 * 10^3 J/kg.°C$

$T_o$ temperature of ice at melting point with value 0°C

$T_i$ is the temperature of ice with value -12°C

$Q_2$ is the energy to move the ice from its its melting point to liquid which is evaluated as

$Q_2 = m_i * L$

Here L is the Latent heat of melting of ice with value $334 * 10^3 J/kg$

$Q_3$ is the energy to move the ice from liquid to the equilibrium temperature which is evaluated as

$Q_1 = m_i * c_w * ( T -T_o)$

$Q_i = m_i [ c_i * ( T_o -T_i) + L + c_w * ( T -T_o) ]$

=> $Q_i = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3 + 4200 * ( 75 - 0) ]$

From

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

We have that

$m_i * 673600 =15750$

=> $m_i = \frac{15750}{673600}$

=> $m_i = 0.0234 \ kg$

8 0

4 years ago