Answers
Answer:
44,640 ft
Explanation:
assuming the rocket started from rest, then v₀ = 0
2 min = 120 s
Δx = v₀t + 1/2at²
Δx = 0 + 1/2(6.2 ft/s²)(120 s)² = 44,640 ft ≈ 8.45 mi
Related Questions
A wire with mass 90.0 g is stretched so that its ends are tied down at points 98.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude of 0.300 cm at the antinodes. Part A What is the speed of propagation of transverse waves in the wire
Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generated by seven-time-winner Lance Armstrong (m = 75.0 kg) is 6.50 W per kilogram of his body mass. (a) How much work does he do during a 85-km race in which his average speed is 10.5 m/s? J (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule = 2.389 10-4 nutritional Calories. nutritional Calories
Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?
Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.
Answers
Answer:
Explanation:
We are given that
Amount of work needed to bring a point charge q0 from infinity to a point P=
We know that potential at point P=
Where U=Amount of work needed to bring a point charge q from infinity to a point P
Initially ,
New charge, q=
Then, work done,U=
Hence, the amount of work needed to bring a new charge 4q0 from infinity to point P=
Answer:
Explanation:
V = u / q,
Work = P = V
1U / 1/4 = 4U
Answers
Answer:
a) W = 25.872 J
b) - 35.28 J
c) - 9.408
Explanation:
a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)
F = - mg (gravity is acting downwards)
F = - 0.6 × 9.8 = - 5.88 N
(H₂ - H₁) = (1.6 - 6) = - 4.4 m
W = (-5.88)(-4.4) = 25.872 J
b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J
c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J
b. tension and gravity
c. tension, gravity, and the centripetal force
d. tension, gravity, the centripetal force, and friction
Answers
Final answer:
The forces directly acting on the ball hanging from a rear-view mirror while a car drives in a circle are tension, gravity, and the centripetal force.
Explanation:
The correct answer is c. tension, gravity, and the centripetal force.
When the car is driving in a circle, the ball experiences both tension and gravity. The tension in the string is what keeps the ball from falling, while gravity pulls the ball downward.
In addition to tension and gravity, the ball also experiences the centripetal force. This force is directed towards the center of the circular motion and keeps the ball moving in a circular path.
Learn more about Forces acting on a ball in circular motion here:
#SPJ12
c) cubic equation
d) a higher order equation
Answers
Answer:
a) Linear equation
Explanation:
Definition of acceleration
if a=constant and we integrate the last equation
So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.
Answers
Answer:
a) W₁ = - 127 J, b) W₂ = 148.18 J, c) = 3.43 m/s and d)
= 3.43 m / s
Explanation:
The work is given the equation
W = F. d
Where the bold indicates vectors, we can also write this expression take the module of each element and the angle between them
W = F d cos θ
They give us displacement, let's use Newton's second law to find strength, like the block has an equal acceleration (a = g / 7). We take a positive sign down as indicated
W-T = m a
T = W -m a
T = mg -mg/7
T = mg 6/7
T = 3.6 9.8 6/7
T = 30.24 N
Now we can apply the work equation to our problem
a) the force of the cord is directed upwards, the displacement is downwards, so there is a 180º angle between the two
W₁ = F d cos θ
W₁ = 30.24 4.2 cos 180
W₁ = - 127 J
b) the force of gravity is directed downwards and the displacement is directed downwards, the angle between the two is zero (T = 0º)
W₂ = (mg) d cos 0º
W₂ = 3.6 9.8 4.2
W₂ = 148.18 J
c) kinetic energy
K = ½ m v²
Let's calculate speed with kinematics
² = vo² + 2 a y
v₀ = 0
a = g / 7
² = 2g / 7 y
= √ (2 9.8 4.2 / 7)
= 3.43 m/s
We calculate
K = ½ 3.6 3.43²
K = 21.18 J
d) the speed of the block and we calculate it in the previous part
= 3.43 m / s
Answers
Answer:
The 1.5V battery can power the flashlight bulb drawing 0.60A for 83.33 minutes before it is depleted.
Explanation:
To determine how long a 1.5V battery can power a flashlight bulb drawing 0.60A, you can use the formula for calculating the energy (in joules) consumed by an electrical device over time:
Energy (Joules) = Power (Watts) × Time (Seconds)
In this case, the power (P) is given by the product of the voltage (V) and current (I):
Power (Watts) = Voltage (Volts) × Current (Amperes)
So, first, calculate the power consumption of the flashlight bulb:
Power (Watts) = 1.5V × 0.60A = 0.90 Watts
Now, you want to find out how long the battery can power the bulb, so rearrange the energy formula to solve for time:
Time (Seconds) = Energy (Joules) / Power (Watts)
Given that the battery stores 4.5 kJ (kilojoules), which is equivalent to 4,500 joules, and the power consumption is 0.90 watts:
Time (Seconds) = 4,500 J / 0.90 W = 5,000 seconds
Now, to express the time in more practical units, convert seconds to minutes:
Time (Minutes) = 5,000 seconds / 60 seconds/minute ≈ 83.33 minutes
So, the 1.5V battery can power the flashlight bulb drawing 0.60A for approximately 83.33 minutes before it is depleted.