What is obliquity in earth's orbit?

Use what you know about mass and how you use it to calculate force in the following situation. If each washer has a mass of 4. 9

mina [271]

Considering the change of units and the definition of weight, the mass of two washers would be 0.0098 kg and if 4 washers are attached to the string, then the applied force on the car will be 0.196 N.

Mass of two washers

If each washer has a mass of 4.9 g, then the mass of two washers will be calculated simply by multiplying the mass of 1 washer by the amount you have of them, in this case 2:

4.9 g×2= 9.8 g

Being 1 kg = 1000 g, you apply the following rule of three to make the change of units: if 1000 g equals 1 kg, 9.8 g equals how many kilograms?

$mass=\frac{9.8 gramsx1 kg}{1000g}$

Solving:

mass= 0.0098 kg

Finally, the mass of two washers would be 0.0098 kg.

Applied force

On the other side, mass is the amount of matter that a body contains and weight is the action exerted by the force of gravity on the body.

The mass of an object will always be the same, no matter where it is located. Instead, the weight of the object will vary according to the force of gravity acting on it.

Weight is calculated as the product of the object's mass and the value of the gravitational acceleration:

W= m×g

In this case, you know:

m= 4.9 g×4= 19.6 g= 0.0196 kg → the mass of 4 washers in kilograms
g= 10 m/s²

Replacing:

W= 0.0196 kg× 10 m/s²

Solving:

W= 0.196 N

Finally, if 4 washers are attached to the string, then the applied force on the car will be 0.196 N.

Learn more:

7 0

3 years ago

Read 2 more answers

A centrifugal pump rotates at n˙ = 740 rpm. Suppose the pump has some swirl at the inlet (α1 = 10°) and exits at an angle of 35°

Blizzard [7]

Answer:

Net head = 380cm

bhp = 17.710kW

Explanation:

Angular velocity of centrifugal pump:

$w=\frac{2\pi n}{60}=\frac{2\pi (750)}{60}=78.54\frac{rad}{s}$

Normal velocity component at outlet of pump:

$V_{2,n}=\frac{V}{2\pi r_{2}b_{2}}=\frac{0.573}{2\pi (0.24)(0.162)}}=2.346\frac{m}{s}$

Tangential velocity component at exit of the pump:

$V_{2,t}=v_{2,n}tan\alpha _{2}=(2.346)tan(35)=1.643\frac{m}{s}$

Normal velocity component at inlet of pump:

$V_{1,n}=\frac{V}{2\pi r_{1} b_{1}}=\frac{0.573}{2\pi (0.12)(0.18)}=4.22\frac{m}{s}$

Tangential velocity component at inlet of the pump:

$V_{1,t}=v_{1,n}tan\alpha _{1}=(4.22)tan(0)=0\frac{m}{s}$

Equivalent head in centimetre of water column:

$H_{water}=H(\frac{rho_{air}}{rho_{water}})\\\\H_{water}=(\frac{w}{g} )(r_{2}V_{2,t}-r_{1}V_{1,t})(\frac{rho_{air}}{rho_{water}}) \\\\H_{water}=(\frac{78.54}{9.81})((0.24)(1.643)-(0.12)(0)})(\frac{1.2}{998})=38m=380cm$

Break horse power:

$bhp=rho_{water} gHV=rho_{water} g[(\frac{w}{g})(r_{2}V_{2,t}-r_{1}V_{1,t})]V\\\\bhp=(998)(9.81)[(\frac{78.54}{9.81})((0.24)(1.643)-(0.12)(0))](0.573)=17710W=17.710kW$

8 0

3 years ago

A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

8 0

3 years ago

What happens to the magnetic field lines if like poles are repelling?

Elza [17]

The north and North Pole or the south and the South Pole are facing each other which is causing them to repel

8 0

3 years ago

Read 2 more answers

All simple machines are variations of which two basic machines?

SCORPION-xisa [38]

Lever; inclined plane

5 0

3 years ago