A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net work done on the object if its velocity changes to 1 8.00 i ^ 1 4.00 j ^

Answers

Answer 1

Answer: (a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= √((6.00m/s)^2+(2.00 m/s)^2)=6.32 m/s$
And so, the kinetic energy of the object is
$K= (1)/(2)mv^2= (1)/(2)(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= √((8.00 m/s)^2+(4.00 m/s)^2)=8.94 m/s$
And so the new kinetic energy is
$K= (1)/(2)mv^2= (1)/(2)(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

Answer 2

Answer:

Final answer:

The initial kinetic energy of the 3.00-kg object traveling at a velocity of 2.00 m/s is 6.00 Joules. When the object's velocity changed to 4.47 m/s, its kinetic energy became 30.02 Joules. Hence, the net work done on the object is 24.02 Joules.

Explanation:

The kinetic energy of any object can be calculated using the formula KE = 0.5 * m * v^2, where m is the object's mass and v is its velocity. For the 3.00-kg object with a velocity of 6.00 i ^ 2 and 2.00 j ^2 m/s, its velocity magnitude would be the square root of (6.00^2 + 2.00^2), which is 2.00 m/s. Plugging the values into the formula, the kinetic energy (a) would be 0.5 * 3.00 * 2.00^2 = 6.00 Joules.

The net work done on an object (b) can be obtained by finding the change in kinetic energy when the object’s velocity changes to 8.00 I and 4.00 j. The final velocity's magnitude would be the square root of (8.00^2 + 4.00^2), which is 4.47 m/s. Hence, the final kinetic energy is 0.5 * 3.00 * 4.47^2 = 30.02 Joules. Therefore, the net work done equals the change in kinetic energy, which is 30.02 - 6.00 = 24.02 Joules.

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On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?

Answers

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

WILL MARK BRAINLIEST PLS HELPPP -- Which of Newton’s Laws explains why the satellite would collide with the moon if gravity is “turned off?”picture attached

Answers

Answer:

Explanation:

B is the answer sorry for the late response

A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? magnitude m/s2 direction (b) What is the velocity of the ball when it reaches its maximum height? magnitude m/s direction (c) Find the initial velocity of the ball. m/s upward (d) Find the maximum height it reaches. m'

Answers

(a) 9.8 m/s^2, downward

There is only one force acting on the ball while it is in flight: the force of gravity, which is

F = mg

where

m is the mass of the ball

g is the gravitational acceleration

According to Newton's second law, the force acting on the ball is equal to the product between the mass of the ball and its acceleration, so

F = mg = ma

which means

a = g

So, the acceleration of the ball during the whole flight is equal to the acceleration of gravity:

g = -9.8 m/s^2

where the negative sign means the direction is downward.

(b) v = 0

Any object thrown upward reaches its maximum height when its velocity is zero:

v = 0

In fact, at that moment, the object's velocity is turning from upward to downward: that means that at that instant, the velocity must be zero.

The initial velocity of the ball can be found by using the equation:

v = u + at

Where

v = 0 is the velocity at the maximum height

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration

t is the time at which the ball reaches the maximum height: this is half of the time it takes for the ball to reach again the starting point of the motion, so

$t=(1.77 s)/(2)=0.89 s$

So we can now solve the equation for u, and we find:

$u=v-at=0-(-9.8 m/s^2)(0.89 s)=8.72 m/s$

(d) 3.88 m

The maximum height reached by the ball can be found by using the equation:

$v^2 - u^2 = 2ad$

where

v = 0 is the velocity at the maximum height

u = 8.72 m/s is the initial velocity

a = g = -9.8 m/s^2 is the gravitational acceleration

d is the maximum height reached

Solving the equation for d, we find

$d=(v^2-u^2)/(2a)=(0^2-(8.72 m/s)^2)/(2(-9.8 m/s^2))=3.88 m$

A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°C. What will be the volume of the air in the bag if you take it with you, still sealed, to the mountains where the temperature is 7.00°C and atmospheric pressure is 70.0 kPa

Answers

The volume of the air in the bag of potato chips to the mountains which is still sealed, 2.766 liters.

What is the gas law?

The gas law is used to show the relationship between the pressure and the temperature of the gases. It can be given as,

$PV=nrT$

Here, (n) and (r) are the constant. Therefore,

$(PV)/(T)=\rm Constant$

For the initial and final values, the gas law can be given as,

$(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$

Here, (subscript 1,and 2) is used for the initial and final amount of pressure and temperature.

The initial values of the bag of potato chips as volume of 2.00 L, pressure of 1.00 ATM and a temperature of 20.0°C. It is known that the value of 1 ATM is equal to the 101.325 kPa.

The final temperature of the pack is 7.00°C and atmospheric pressure is 70.0 kPa

Put the values in the above formula as,

$(101.325*2)/(293)=(70* V_2)/(280)\nV_2=2.766\rm \; liters$

Hence, the volume of the air in the bag of potato chips to the mountains which is still sealed, 2.766 liters.

Learn more about the gas law here;

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Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_(1) = 2.00\ L$

Pressure $P_(1)= 1.00\ atm$

Pressure $P_(2)= 70.0\ kPa$

Temperature $T_(1)= 20.0°C = 293\ K$

Temperature $T_(2)= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$(PV)/(T)=\ Constant$

For both temperature,

$(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))$

Put the value into the formula

$(101.325*2)/(293)=(70* V_(2))/(280)$

$V_(2)=(101.325*2*280)/(293*70)$

$V_(2)=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a wavelength of 3.5 cm?Express your answer in GHz and using two significant figures.
f = ________GHz
Microwave signals are beamed between two mountaintops 52 km apart. How long does it take a signal to travel from one mountaintop to the other?
Express your answer in ms and using two significant figures.
t = ________ms

Answers

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =(c)/(\lambda) =(3e8m/s)/(0.035m) = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =(d)/(t) (3)$

where v= c= speed of light in vacuum = 3*10⁸ m/s

d= distance between mountaintops = 52 km = 52*10³ m

Solving for t, we get:

$t = (d)/(c) = (52e3m)/(3e8m/s) = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)$

⇒ t = 0.2 ms (with two significative figures)

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s , the induced emf in the second coil has a magnitude of 1.60×10−3 V . Part A What is the mutual inductance of the pair of coils? MM = nothing H Request Answer Part B If the second coil has 22 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ? ΦΦ = nothing Wb