Which three of the following are characteristics of ionic bonding?

What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem

8090 [49]

Answer:

$\Delta L = 4.2 m$

Explanation:

As per the formula of thermal expansion we know that

$L = L_o(1 + \alpha\Delta T)$

so here we will have

$L_o = 1400 m$

$\alpha = 12 \times 10^{-6} per ^oC$

$\Delta T = 250 degree C$

so here change in the length of the rod is given as

$\Delta L = L - L_o$

$\Delta L = L_o \alpha \Delta T$

$\Delta L = 1400 (12 \times 10^{-6})(250)$

$\Delta L = 4.2 m$

8 0

3 years ago

When an object moves in uniform circular motion, the direction of its acceleration is?

Harrizon [31]

Clock wise idk i think you should double check my answer

7 0

2 years ago

What is the acceleration of a race car if it has a mass of 1200kg and is moving with an engine force of 400N

meriva

Answer:

0.34 m/s^2

Explanation:

force=mass × acceleration

400 =1200 × acceleration

acceleration=400/1200

=0.34 m/s^2

5 0

2 years ago

A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl

guajiro [1.7K]

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

$P_1 = m_av_a + m_b v_b$

$P_1 = 1300 \times 35+ 1000 \times 25$

$P_1 =70500 Kg.m/s$

final momentum

$P_2 = m_aV_a + m_b V_b$

$P_2 = 1300 \times 30+ 1000 \times 31.5$

$P_2 =70500 Kg.m/s$

here initial momentum is equal to the final momentum of the car.

hence, momentum is conserved in the collision.

6 0

2 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

$p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}$

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

$F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N$

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

$F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}$

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0

3 years ago