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Setler [38]
2 years ago
7

Which of the following correctly describes the law of conservation of matter?

Physics
1 answer:
Tems11 [23]2 years ago
7 0

Answer:

C

Explanation:

Tama Po yan don't worry

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Explain the relationship between speed, velocity and acceleration
Montano1993 [528]
Speed is the rate at which something covers a distance; velocity is the same but it takes into account whether it goes forwards or backwards; and acceleration is the rate of an increase in speed.
8 0
3 years ago
Relative motion can best be defined as
geniusboy [140]
Relative motion can best be defined as B<span> the motion of one object as it appears to another object.
An example is when you are in a car the car has the actual motion because it is the one moving but you are also moving because of relative motion.</span>
6 0
2 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
2 years ago
an engine has been design to work between source and the sink at temperature 177 degree Celsius and 27 degree Celsius respective
irina [24]

<u>Given data</u>

Source temperature (T₁) = 177°C = 177+273 = 450 K

Sink temperature (T₂) = 27°C = 27+273 = 300 K

Energy input (Q₁) = 3600 J ,

Work done = ?

                We know that, efficiency (η) = Net work done ÷ Heat supplied

                                                           η =   W ÷ Q₁  

                                                           W = η × Q₁

               First determine the efficiency ( η ) = ?

                                Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)

                                                                        = 33.3% = 0.333

               Now, Work done is W = η × Q₁

                                                    = 0.33 × 3600

                                                 <em>  W = 1188 J</em>

<em>Work done by the engine is 1188 J</em>

4 0
3 years ago
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