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2 years ago

Which of the following elements are most likely to have similar chemical properties?

Physics

1 answer:

NNADVOKAT [17]2 years ago

5 0

Answer:

transition metal, and inner transition metals groups are numbered 1-18 from left to right

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Unit 5 lesson 7 physical science 12 question

Kryger [21]

I just took it 100% 11/11
1.D
2.A
3.A
4.A
5.B
6.C
7.D
8C
9A
10B
11C

8 0

3 years ago

9- Under what circumstances would a vector have components that are equal in

valkas [14]

Explanation:

c. if the vector is oriented at 0° from the X -axis.

6 0

2 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0

2 years ago

A Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does

Gelneren [198K]

Answer:

D. 130 J

Explanation:

The coefficient of performance for a machine that is being used to cool, is given by:

$COP=\frac{Q_C}{W}=\frac{T_C}{T_H-T_C}$

Here $Q_C$ is the heat removed from the cold reservoir, W is the work required, that is, the energy required to remove the heat from the interior of the house, $T_C$ is the cold temperature and $T_H$ is the hot temperature. Recall use absolutes temperatures( $273.15+^\circ C$ ). Replacing and solving for W:

$W=Q_c\frac{T_H-T_C}{T_C}\\W=2000J\frac{312.15K-293.15K}{293.15K}\\W=129.63J$

8 0

3 years ago

Why is understanding the concept of forces, friction and gravity important?

nydimaria [60]

Because it's the basis of how everything around you works

8 0

3 years ago