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2 years ago

The fact that some asteroids cluster in what are called asteroid families is probably the result of:

Physics

1 answer:

alisha [4.7K]2 years ago

8 0

Answer:

The families are thought to form as a result of collisions between asteroids. In many or most cases the parent body was shattered, but there are also several families which resulted from a large cratering event which did not disrupt the parent body (e.g. the Vesta, Pallas, Hygiea, and Massalia families).

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A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the

Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

$W_{r} = |F_{r}|*|d|*cos(\theta_{1})$

Where:

$F_{r}$ : is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

$W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J$

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

$W_{g} = |F_{g}|*|d|*cos(\theta_{2})$

The force of gravity is given by the weight of the bike.

$F_{g} = -mgsin(24)$

And the angle between the force of gravity and the direction of motion is 180°.

$W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})$

$W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J$

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.

I hope it helps you!

3 0

3 years ago

On another planet gravity has a value of 5.5 m/s . If an object is dropped how long will it take to fall 53 m?

Nikolay [14]

Answer:

4.4 seconds

Explanation:

Given:

a = -5.5 m/s²

v₀ = 0 m/s

y₀ = 53 m

y = 0 m

Find: t

y = y₀ + v₀ t + ½ at²

0 = 53 + 0 + ½ (-5.5) t²

0 = 53 − 2.75 t²

t = 4.39

Rounded to two significant figures, it takes 4.4 seconds for the object to land.

7 0

3 years ago

Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3

VARVARA [1.3K]

Answer:

Part a)

$f = 4.76 \times 10^{14} Hz$

Part b)

$d = 3.48 \times 10^{-4} m$

Part c)

$\theta = 0.311 degree$

Explanation:

Part a)

As we know that the speed of light is given as

$c = 3 \times 10^8 m/s$

$\lambda = 630 nm$

now the frequency of the light is given as

$f = \frac{c}{\lambda}$

so we have

$f = \frac{3 \times 10^8}{630 \times 10^{-9}}$

$f = 4.76 \times 10^{14} Hz$

Part b)

Position of Nth maximum intensity on the screen is given as

$y_n = \frac{n\lambda L}{d}$

so here we know for 3rd order maximum intensity

$y_3 = 0.76 cm$

n = 3

L = 1.4 m

$0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}$

$d = 3.48 \times 10^{-4} m$

Part c)

angle of third order maximum is given as

$d sin\theta = 3 \lambda$

$3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})$

$\theta = 0.311 degree$

8 0

3 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

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5 0

3 years ago

We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o

Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

Learn more about the acceleration here:

brainly.com/question/22048837

#SPJ10

6 0

2 years ago