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frosja888 [35]
3 years ago
8

Which element’s symbol could replace the question mark in the diagram? carbon (C) tellurium (Te) beryllium (Be) cesium (Cs)

Physics
1 answer:
ioda3 years ago
8 0

Answer:

Tellurium

Explanation:

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If a typical smelt weighs 225 g, what is the total mass of pcbs in a smelt in the great lakes? If a typical smelt weighs 225 g,
photoshop1234 [79]

Answer;

= 2.34 x 10^-4 g pcbs

Explanation;

-A typical smelt has 1.04 ppm of PCB is great lakes.

Therefore;

The amount of PCB in a smelt in great leak can be calculated by multiplying the mass of smelt by the pcbs present in a smelt

= 225 g * ( 1.04/100000)

= 2.34 x10^-4 g pcbs

7 0
3 years ago
Suppose a car is traveling at +25.0 m/s, and the driver sees a traffic light turn red. After 0.340 s has elapsed (the reaction t
scoundrel [369]
First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m

Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration 
d = <span> 1/2*(625) /(7) = 44.6428 m

The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m
3 0
3 years ago
A pump lifts 400 kg of water per hour a height of 4.5 m .
nasty-shy [4]

Answer:

Power = Work / Time

P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts

Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp

6 0
3 years ago
A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in
miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is \mu_s =  0.60

The value for static friction is \mu_k =  0.70

Explanation:

From the question we are told that

The mass of the clock is m  =  95 \  kg

The first horizontal force is F _s  =  560 \  N

    The second horizontal force is    F _k  =  650  \  N

Generally the static frictional force is equal to the first  horizontal force

So

     F _s  =  m  *  g  *  \mu_s

=>   560  =  95  *  9.8  *  \mu_s

=>    \mu_s =  0.60

Generally the kinetic frictional force is equal to the second horizontal force

So

      F _k  =  m  *  g  *  \mu_k

      650 =  95  *  9.8  *  \mu_k

     \mu_k =  0.70

3 0
3 years ago
PLEASE HELP!!!! Scientists launch a rocket, and they monitor its acceleration and the force exerted by its engines. As the rocke
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4 years ago
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