A 1000 kg car can accelerate from rest to a speed of 25 m/s in 10 s. What average power must the engine of car produce in order to cause this acceleration? Neglect friction losses
What is the gravitational force between two masses of 15kg each, when their centers are 0.25m? Could you detect this force with even sensitive equipment?
If a force of 10 n is applied to an object with a mass of 1kg the object will accelerate at
When exiting the highway, a 1100-kg car is traveling at 22 m/s. The car's kinetic energy decreases by 1.4×105J The exit's speed limit is 35 mi/h. Did the driver reduce its speed enough?
Answer: The final velocity is 33.78 mi/h, so the driver did reduced his speed enough.
Explanation: The kinetic energy of an object can be calculated as:
K = (1/2)m*v^2
We know that the mass of the car is m=1100kg
and the initial velocity is 22m/s
The initial kinetic energy is:
K = (1/2)*1100*(22)^2 = 266,200 joules.
Now, if the kinetic energy decreases by 1.4x10^5 J, the new kinetic energy is:
K = 266,200j - 140,000j = 126,200j
So we now can find the new velocity in m/s.
126,200 = (1/2)*1100*v^2
126,200*2/1100 = v^2
229.45 = v^2
v = (229.45)^(1/2) = 15.1 m/s
We know that the limit is 35 mi/h, so we need to transform our result into miles per hour.
We know that in one hour, there are 3600 seconds, so the velocity per hour is:
15.1*3600 m/h = 54,360 m/h
and we know that one mile is 1609.34 meters, so we need to divide by 1609.34.
v = (54,360/1609.34) mi/h = 33.78 mi/h
this is less than the speed limit, so the driver reduced his speed enough.
After losing kinetic energy, the car's final velocity was approximately 18.99 m/s, exceeding the exit's speed limit of 15.64 m/s, hence the driver did not reduce their speed adequately.
Explanation:To determine whether the driver reduced their speed enough when exiting the highway, we must calculate the car's speed after its kinetic energy decreases by 1.4×105J. The initial kinetic energy (KE) of the 1100-kg car traveling at 22 m/s can be calculated using the equation KE = ½ mv². Plugging in the values, we can find the initial kinetic energy:
KEinitial = ½ (1100 kg)(22 m/s)² = 5.28×105J
After losing 1.4×105J of energy, the remaining kinetic energy will be:
KEfinal = KEinitial - 1.4×105J = (5.28 - 1.4)×105J = 3.88×105J
We can solve for the final velocity (vfinal) using the remaining kinetic energy:
½ (1100 kg)vfinal² = 3.88×105J
vfinal = √((2×3.88×105J) / 1100 kg)
vfinal ≈ 18.99 m/s
To compare to the speed limit, we convert 35 mi/h to meters per second:
35 mi/h × 0.44704 (conversion factor) = 15.64 m/s
Since the final velocity of the car is 18.99 m/s, which is greater than the exit's speed limit of 15.64 m/s, the driver did not reduce their speed enough.
A ball is dropped from a height of 100 ft. one second later another ball is dropped from a height of 75 ft, which one hits the ground first? ...?
Both balls will hit the ground simultaneously because the time it takes to fall is based on the height, not the initial velocity.
Explanation:The time it takes for an object to fall to the ground depends only on its height and is independent of its initial velocity. In this case, both balls are dropped, so they have an initial velocity of 0 m/s. The ball dropped from 100 ft and the ball dropped from 75 ft will both hit the ground at the same time. This is because the only factor affecting the time it takes to fall is the height, not the initial velocity. Therefore, both balls will hit the ground simultaneously.
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A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 0.700 revolutions per second. What is the centripetal acceleration of the ball?
The centripetal acceleration of the ball will be 38.68 meter per sq.second.
What is centripetal acceleration?The acceleration needed to move a body in a curved way is understood as centripetal acceleration.
The direction of centripetal acceleration is always in the path of the center of the course.
The given data in the problem;
r is the radius= 2.00m
frequency (f) = 0.7 rev/s
The angular velocity is found as;
[tex]\rm \omega = 2 \pi f \\\\ \omega = 2 \times 3.14 \times 0.7 \\\\ \omega = 4.398 \ rad/sec[/tex]
The centripetal acceleration is given by;
[tex]\rm a_c= \omega^2r \\\\\ a_c= (4.398)^2 \times 2.00 \\\\ a_c=38.68 \ m/sec^2[/tex]
Hence, the centripetal acceleration of the ball will be 38.68 meter per sq.second.
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