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A small cannonball with mass 9 kilograms is shot vertically upward with an initial velocity of...
Submission 3 (0.66/1 points) Friday, July 24, 2020 07:15 PM +08 Suppose a small cannonball weighing...
Submission 3 (0.66/1 points) Friday, July 24, 2020 07:15 PM +08 Suppose a small cannonball weighing 98 N is shot vertically upward, with an initial velocity Vo = 94 m/s. The answer to the question "How high does the cannonball go?" depends on whether we take air resistance into account. If air resistance is ignored and the positive direction is upward, then a model for the state of the cannonball is given by d s/dt = -9 (equation (12) of...
P1.A cannonball is shot at an angle of 45" with an initial velocity vo. This is...
P1.A cannonball is shot at an angle of 45" with an initial velocity vo. This is calculated to exactly land at point B in a projectile trajectory. However, the people at point B launched a rocket with an initial ve locity at 20 from the horizontal sometime after the cannonball launched. The rocket, which will intercept the cannonball at some vertical distance from the ground level, is in a linear path, Find (a) horizontal distance between point A and Point...
2) (15PTS) A BODY of Mass in FALLING VERTICALLY IN SPACE ENCOUNTERS AIR RESISTANCE PROPORTIONAL TO...
2) (15PTS) A BODY of Mass in FALLING VERTICALLY IN SPACE ENCOUNTERS AIR RESISTANCE PROPORTIONAL TO THE StU ARE OF ITS INSTANTANEOUS VELOCITY vlt) in meters/sec. ITS DIFFERENTIAL EQUATION OF MOTION IS m du = mg - kv²; vco)= Vo where Kyo is THE CONSTANT OF PRPORT, ON ALITY AND J is POSITIVE. FIND THE TERMINAL VELOCITY OF THE FALLING BODY ( t o )
(1 point) An object of mass 5 kg is given an upward initial velocity of 16...
(1 point) An object of mass 5 kg is given an upward initial velocity of 16 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -50, where v is the velocity of the object in m/sec. Assume gravitational constant is g = 9.8m/seca. Set up the differential equation for this scenario: v' = m/sec Solve the differential equation for the equation of motion: The equation is both...
An object of mass m 5 kilograms falls vertically to the ground under the action of...
An object of mass m 5 kilograms falls vertically to the ground under the action of the earth gravitational acceleration of magnitude g 10 meters per second squared. Denote by y vertical coordinate, positive upwards, and let y 0 be at the earth surface. Recall that the force on the object in this situation is f--mg, where the negative sign says the force points downwards. (a) Write the differential equation satisfied by this system. y"-10 Note: Write t for t,...
(1 point) A water balloon of mass 380 grams is launched with an initial (horizontal) velocity...
(1 point) A water balloon of mass 380 grams is launched with an initial (horizontal) velocity of 43 meters per second. As it travels, water leaks from the balloon at a rate of 60 grams per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second. Again, because the motion is horizontal, ignore any effect due to gravity (a) Find the velocity of the balloon as a function of time. v(t) (b) What is the furthest...
(1 point) A water balloon of mass 380 grams is launched with an initial (vertical) velocity...
(1 point) A water balloon of mass 380 grams is launched with an initial (vertical) velocity of 43 meters per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second, and use 9.81 meters per second squared for the acceleration due to gravity. (a) Find the height of the balloon as a function of time. h(t) = (b) What is the terminal velocity of the balloon? (Enter your answer as a positive velocity.) The terminal velocity...
(1 point) A water balloon of mass 380 grams is launched with an initial (horizontal) velocity...
(1 point) A water balloon of mass 380 grams is launched with an initial (horizontal) velocity of 43 meters per second. As it travels, water leaks from the balloon at a rate of 60 grams per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second. Again, because the motion is horizontal, ignore any effect due to gravity (a) Find the velocity of the balloon as a function of time. v(t) (b) What is the furthest...
(2 points) A body of mass 4 kg is projected vertically upward with an initial velocity 40 meters ...
(2 points) A body of mass 4 kg is projected vertically upward with an initial velocity 40 meters per second. The gravitational constant is g-9.8m/82. The air resistance is equal to kļul where k is a constant. Find a formula for the velocity at any time ( in terms of k ): Find the limit of this velocity for a fixed time t0 as the air resistance coefficient k goes to O. (Enter t0 as tzero.) (t0) How does this...
PLEASE ANSWER COMPLETELY A projectile of mass m launched vertically upward from the origin (0,0) with...
PLEASE ANSWER COMPLETELY
A projectile of mass m launched vertically upward from the origin (0,0) with an initial velocity of vo explodes into three pieces when it reaches its maximum height of 490.0 m. Immediately after the explosion, one piece with mass m/4 is observed to have velocity v 3vot vo/2 y, and a second piece also of mass m/4 is observed to have velocity v Vox (a) What is the velocity v of the third piece immediately after the...
Submission 3 (0.66/1 points) Friday, July 24, 2020 07:15 PM +08 Suppose a small cannonball weighing 98 N is shot vertically upward, with an initial velocity Vo = 94 m/s. The answer to the question "How high does the cannonball go?" depends on whether we take air resistance into account. If air resistance is ignored and the positive direction is upward, then a model for the state of the cannonball is given by d s/dt = -9 (equation (12) of...
P1.A cannonball is shot at an angle of 45" with an initial velocity vo. This is calculated to exactly land at point B in a projectile trajectory. However, the people at point B launched a rocket with an initial ve locity at 20 from the horizontal sometime after the cannonball launched. The rocket, which will intercept the cannonball at some vertical distance from the ground level, is in a linear path, Find (a) horizontal distance between point A and Point...
2) (15PTS) A BODY of Mass in FALLING VERTICALLY IN SPACE ENCOUNTERS AIR RESISTANCE PROPORTIONAL TO THE StU ARE OF ITS INSTANTANEOUS VELOCITY vlt) in meters/sec. ITS DIFFERENTIAL EQUATION OF MOTION IS m du = mg - kv²; vco)= Vo where Kyo is THE CONSTANT OF PRPORT, ON ALITY AND J is POSITIVE. FIND THE TERMINAL VELOCITY OF THE FALLING BODY ( t o )
(1 point) An object of mass 5 kg is given an upward initial velocity of 16 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -50, where v is the velocity of the object in m/sec. Assume gravitational constant is g = 9.8m/seca. Set up the differential equation for this scenario: v' = m/sec Solve the differential equation for the equation of motion: The equation is both...
An object of mass m 5 kilograms falls vertically to the ground under the action of the earth gravitational acceleration of magnitude g 10 meters per second squared. Denote by y vertical coordinate, positive upwards, and let y 0 be at the earth surface. Recall that the force on the object in this situation is f--mg, where the negative sign says the force points downwards. (a) Write the differential equation satisfied by this system. y"-10 Note: Write t for t,...
(1 point) A water balloon of mass 380 grams is launched with an initial (horizontal) velocity of 43 meters per second. As it travels, water leaks from the balloon at a rate of 60 grams per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second. Again, because the motion is horizontal, ignore any effect due to gravity (a) Find the velocity of the balloon as a function of time. v(t) (b) What is the furthest...
(1 point) A water balloon of mass 380 grams is launched with an initial (vertical) velocity of 43 meters per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second, and use 9.81 meters per second squared for the acceleration due to gravity. (a) Find the height of the balloon as a function of time. h(t) = (b) What is the terminal velocity of the balloon? (Enter your answer as a positive velocity.) The terminal velocity...
(1 point) A water balloon of mass 380 grams is launched with an initial (horizontal) velocity of 43 meters per second. As it travels, water leaks from the balloon at a rate of 60 grams per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second. Again, because the motion is horizontal, ignore any effect due to gravity (a) Find the velocity of the balloon as a function of time. v(t) (b) What is the furthest...
(2 points) A body of mass 4 kg is projected vertically upward with an initial velocity 40 meters per second. The gravitational constant is g-9.8m/82. The air resistance is equal to kļul where k is a constant. Find a formula for the velocity at any time ( in terms of k ): Find the limit of this velocity for a fixed time t0 as the air resistance coefficient k goes to O. (Enter t0 as tzero.) (t0) How does this...
PLEASE ANSWER COMPLETELY
A projectile of mass m launched vertically upward from the origin (0,0) with an initial velocity of vo explodes into three pieces when it reaches its maximum height of 490.0 m. Immediately after the explosion, one piece with mass m/4 is observed to have velocity v 3vot vo/2 y, and a second piece also of mass m/4 is observed to have velocity v Vox (a) What is the velocity v of the third piece immediately after the...
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