Anyone know the answer

The elements that are considered very reactive are part of which family

s344n2d4d5 [400]

#ClutchMoments lol !

4 0

4 years ago

In which parking situations should you use your parking brake?

VMariaS [17]

Answer:
-You should use your parking brake when on a hill/mountain with a huge drop because your car can’t slide off the edge.
-When there is a flood where your car is sitting because the water can’t carry your car

5 0

4 years ago

Read 2 more answers

A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H

Reika [66]

Answer:

The answer to your question is:

a) t1 = 2.99 s ≈ 3 s

b) vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ? time to reach the ground

vf = ? final speed

a) h = vot + (1/2)gt²

74 = 10t + (1/2)9.81t²

4.9t² + 10t -74 = 0 solve by using quadratic formula

t = (-b ± √ (b² -4ac) / 2a

t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

t = (-10 ± √ 1550.4 ) / 9.81

t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81

t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81

t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are

no negative times.

b) Formula vf = vo + gt

vf = 10 + (9.81)(3)

vf = 10 + 29.43

vf = 39.43 m/s

4 0

3 years ago

A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is us

Nataliya [291]

Answer:

$T_C$ =118.8 K= 154.2°C

Explanation:

COP_max of carnot heat pump= $\frac{T_{H} }{T_{H}-T_{C} }$

where T_H and T_C are temperatures of hot and cold reservoirs

Also COP= $\frac{Q_H}{W}$

in the question $COP= \frac{60}{100} \times COP_{max}$

⇒ $\frac{Q_H}{W} =\frac{60}{100}\times\frac{T_H}{T_H-T_C}$

heat is added directly to be as efficient as via heat pump

$Q_H= W$

and T_H= 24° C= 297 K

$1=\frac{60}{100}\times \frac{297}{297-T_C}$

on calculating the above equation we get

$T_C$ =118.8 K

the outdoor temperature for efficient addition of heat to interior of home

$T_C$ =118.8 K= 154.2°C

6 0

3 years ago

A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to

Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV

V = W/q ....(1)

Using equation (1), the potential difference between points A and B is:

$V_{1}=\frac{W_{1} }{q}$

Substitute the suitable values in the above equation.

$V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }$

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

$V_{2}=\frac{W_{2} }{q}$

Substitute the suitable values in the above equation.

$V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }$

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0

3 years ago