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Sergio [31]
4 years ago
9

Which of the following is Not true about the noble gases

Physics
1 answer:
Alexandra [31]4 years ago
3 0
Noble gases are not highly reactive
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When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t
choli [55]

Answer:

Part a)

\tau = 23.1 Nm

Part b)

\tau = 17.05 Foot pound force

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

\tau = \vec r \times \vec F

now we have

\tau = (0.140)(165)

\tau = 23.1 Nm

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

\tau = 23.1 Nm

\tau = 23.1 (0.225 Lb)(3.28 foot)

\tau = 17.05 Foot pound force

3 0
3 years ago
The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of
motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

F\propto \dfrac{mgv^2}{r}

F=\dfrac{kmgv^2}{r}

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

640=\dfrac{k(2600)(40)^2}{650}

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

F=\dfrac{0.1\times 2600\times (30)^2}{750}

F = 312 N

Hence, this is the required solution.

6 0
3 years ago
Given a force of 100 N and acceleration of 5 m/s2, what is the mass
tatyana61 [14]

Answer:

20 kg

Explanation:

remember the equation f=ma.

100 N=force

5 m/s2= acceleration

so you need to divide force by acceleration: 100 N/ 5 m/s2= 20 kg, to get the mass.

8 0
2 years ago
A small 25 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe’s kinetic energy?
Helen [10]
Kinetic energy = 1/2mv^2

=1/2(25)(1^2)
= 12.5J
4 0
3 years ago
At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v
Cerrena [4.2K]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

R= 6370*10^3 m

v = 239m/s

a = 16.5m/s^2

The circumference of the earth would be

\phi = 2\pi R

Velocity is defined as,

v = \frac{x}{t}

t = \frac{x}{v}

Here x = \phi, then

t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}

t = 167463.97s

Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds

4 0
3 years ago
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