Given the dx and dy = 500 m, and that the rocket is moving at an angle of 30, find initial velocity of x. The answer is 46 m/s, and the time is 12.49 s. Please help!
Homework Answers
Horizontal distance D =Vix*t
D =(Vi*cos theta)*t
Vi =D/(t*cos theta)
Vi =500 m/(12.49 s*cos 30)
Vi=46.225 m/s
Vi =46 m/s
Interpret the rocket equation dv(t)M(t)=-udM(t) [EQ.1] within the framework of the law of momentum conservation, written in...
Interpret the rocket equation dv(t)M(t)=-udM(t) [EQ.1] within the framework of the law of momentum conservation, written in a closed system; here M(t) is the rocket mass, at time t, whereas dM(t) isby definition, dM(t)=M(t+dt)-M(t); -dM(t)=|dM(t)|, is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)=v(t+dt)–v(t), i.e. theincrease in the velocity of the rocket through the period of time dt; u is the relative...
please solve 2 QUESTION 2 a) (5 p) Interpret the rocket equation dv(t)M(1)=-udMO [EQ.1) within the...
please solve 2
QUESTION 2 a) (5 p) Interpret the rocket equation dv(t)M(1)=-udMO [EQ.1) within the framework of the law of momentan conservation, written in a closed system here M(t) is the rocket mass, at time t, whereas dMt) is by definition, dM(t)-M(t+dt)-M(t): -SM(t)-M(!), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is still by definition, dy(t){t+dt)-v(t), i.e. the increase in the velocity of the...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO [EQ.1) within the framework of the...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO [EQ.1) within the framework of the law of momentum conservation, written in a closed system, here Mt) is the rocket mass, time t, whereas M(t) is by definition, dM(t)=M(t+dt)-M(t): -dM(t)-dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt: on the other hand, dv(t) is, still by definition, dv(t)v(t+dt)-vít), i.e. the increase in the velocity of the rocket through the period...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO (EQ.1) within framework of the law...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO (EQ.1) within framework of the law of momentum conservation, written in a closed system, here M(t) is the rocket mass, at time t, whereas M(t) is by definition, dM(t)=M(t+dt)-M(t): -dM(t)=dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)=v(t+dt)-vít).i.e. the increase in the velocity of the rocket through the period of...
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A 16-gram bullet moving at 1825 m/s plunges into 2.5 kg of paraffin wax. The wax was initially at 30°C. Assuming that all the bullet's energy heats the wax, what is its final temperature (in ºC)? Take the mechanical equivalent of heat to be 4 J/cal and the specific heat of wax to be 0.7 cal/g °C. Round your answer to the nearest 2 decimal places.
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Please report your answer to 2 significant figures. As shown below, a block of mass m = 0.37 kg is initially at rest on a frictionless inclined plane at height = 5m and pressed against a spring so that the spring is compressed by an amount x = 2.7 m. Using conservation of mechanical energy, please find the speed of the block after it has lost contact with the spring and has reached a height he = 14 m. The...
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a) (5 p) Interpret the rocker equation dv(t)M(t)=-udM(t) (EQ.1) within the framework of the law of...
a) (5 p) Interpret the rocker equation dv(t)M(t)=-udM(t) (EQ.1) within the framework of the law of momentum conservation, written in a closed system, here M(t) is the rocker mass, at time t, whereas M(t) is by definition, dM(t)-M(t+dt)-M(t): - dM(t)-dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)-v(t+dt)-v(t), i.e. the increase in the velocity of the rocker through the period...
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Task 10 A spring is 3.0 cm and has a spring constant of 2000 N/m. The spring shoots out a bullet with a mass of 0.049 kg. The bullet collides elastically with a very similar bullet hanging in a cord with length L=0.60 m which is attached in a point A. See figure. spring a) About how much speed does the bullet get from the spring? 1) 3.0 m/s 11) 4.2 m/s III) 5.0 m/s IV) 6.1 m/s V) 11...
7.1- WORK: SCIENTIFIC DEFINITION - PROBLEM 1 Example 7.1 Mr. Clean 500 N Figure 7.5 (Example...
7.1- WORK: SCIENTIFIC DEFINITION - PROBLEM 1 Example 7.1 Mr. Clean 500 N Figure 7.5 (Example 7.1) A vacuum cleaner being pulled at an angle of 50.0" from the horizontal A man cleaning a floor pulls a vacuum cleaner with a force of magnitude - 50.0 N at an angle of 30.0 with the horizontal (Fig. 7.5). Calculate the work done by the force on the vacuum cleaner as the vac uum cleaner is displaced 3.00 m to the right...
please solve 2
QUESTION 2 a) (5 p) Interpret the rocket equation dv(t)M(1)=-udMO [EQ.1) within the framework of the law of momentan conservation, written in a closed system here M(t) is the rocket mass, at time t, whereas dMt) is by definition, dM(t)-M(t+dt)-M(t): -SM(t)-M(!), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is still by definition, dy(t){t+dt)-v(t), i.e. the increase in the velocity of the...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO [EQ.1) within the framework of the law of momentum conservation, written in a closed system, here Mt) is the rocket mass, time t, whereas M(t) is by definition, dM(t)=M(t+dt)-M(t): -dM(t)-dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt: on the other hand, dv(t) is, still by definition, dv(t)v(t+dt)-vít), i.e. the increase in the velocity of the rocket through the period...
QUESTION 2 a) (5 p) Interpret the rocket equation dv(OM(t)=-udMO (EQ.1) within framework of the law of momentum conservation, written in a closed system, here M(t) is the rocket mass, at time t, whereas M(t) is by definition, dM(t)=M(t+dt)-M(t): -dM(t)=dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)=v(t+dt)-vít).i.e. the increase in the velocity of the rocket through the period of...
Please report your answer to 2 significant figures. As shown below, a block of mass m = 0.37 kg is initially at rest on a frictionless inclined plane at height = 5m and pressed against a spring so that the spring is compressed by an amount x = 2.7 m. Using conservation of mechanical energy, please find the speed of the block after it has lost contact with the spring and has reached a height he = 14 m. The...
2. (20) k = 500 N/m fo 20 kg The 20 kg block slides on a level, frictionless surface. When the spring is at rest, x=0. a. (10) For F = 0, the spring initially at rest, and the block moving to the right at 3 m/s at t=0, use conservation of energy to find the velocity of the block after it has travelled .4 m to the right. b. (10) Find the differential equation for x and the solution...
a) (5 p) Interpret the rocker equation dv(t)M(t)=-udM(t) (EQ.1) within the framework of the law of momentum conservation, written in a closed system, here M(t) is the rocker mass, at time t, whereas M(t) is by definition, dM(t)-M(t+dt)-M(t): - dM(t)-dM(t), is the mass of the gas thrown by the rocket through the infinitely small period of time dt; on the other hand, dv(t) is, still by definition, dv(t)-v(t+dt)-v(t), i.e. the increase in the velocity of the rocker through the period...
Task 10 A spring is 3.0 cm and has a spring constant of 2000 N/m. The spring shoots out a bullet with a mass of 0.049 kg. The bullet collides elastically with a very similar bullet hanging in a cord with length L=0.60 m which is attached in a point A. See figure. spring a) About how much speed does the bullet get from the spring? 1) 3.0 m/s 11) 4.2 m/s III) 5.0 m/s IV) 6.1 m/s V) 11...
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