A man covers 2.5km in 5 minutes. what will be the speed in km/ hr?
Time is 5 minutes I.e 5/60 = 1/12 hrs Speed = distance / time = 2.5/(1/12) = 2.5 * 12 = 30 km/hr
A constant force F = (1.25i + 4.35j) N acts on an object as it moves along a straight line path. If the obje?
You can calculate the F by any method you chose. This question is simply there to show you two different ways of getting to the same answers. (Both answers are the same) You can solve for B by the following (F_x*d_x)+(F_y*d_y) or (1.25*1.35)+(4.35*4.25) That answer will work for both A and B. Unfortunately, I do not understand the alternate method of solving they are suggesting.
How do police officers check the speed of cars on a highway?
Law enforcement agencies have vehicle-mounted, handheld, and static radar devices that can measure the speed of a moving vehicle. All three types of devices uses Doppler radar; they emit a radio signal in a narrow beam, and the signal bounces of an object, and is reflected back to the device. When the radio signal returns to the device, the frequency has changed, and the device calculates the speed of the object based off of how much the frequency has changed. Handheld devices are considered "stationary tools", which means that the officer who is using the device has to stay where he/she is at. Some vehicles are mounted with stationary radar as well, and some are mounted with "moving" radar, in which the device emits a signal both in front and towards the rear of the vehicle, and so the officer can read the speed of a vehicle while on the move. Some devices are also placed on the side of the road, where you'll see a sing that says "Your speed is ##."
An atomic nucleus of mass m traveling with speed v collides elastically with a target particle of mass 2m?
Conserve momentum! In original direction: p = mv = 2mUcosΘ where U is the post-collision velocity of the 2m particle and Θ is measured relative to the original direction of the m particle. → Then v = 2UcosΘ In transverse direction: p = 0 = 2mUsinΘ - mV where V is the post-collision velocity of the m particle, now moving at 90º to its original direction. → Then V = 2UsinΘ Divide second by first: V / v = 2UsinΘ / 2UcosΘ = tanΘ → V = v*tanΘ For an elastic collision, we can also conserve energy: ½mv² = ½mV² + ½(2m)U² → m/2 cancels, leaving v² = V² + 2U² → substitute for V and v (2UcosΘ)² = (2UsinΘ)² + 2U² 4U²cos²Θ = 4U²sin²Θ + 2U² → 2U² cancels, leaving 2cos²Θ = 2sin²Θ + 1 I know of only two ways to solve this; iteration and wolfram. wolfram gives Θ = π/6 = 30º Then V = vtan30 = 0.577v = v / √3 = v√3 / 3 and U = v / 2cos30 = v / (2*√3/2) = v / √3 = 0.577v They have equal speeds. Since they have the same speeds and the target particle has twice the mass, then 2/3 of the original KE was transferred to the target particle.
A motorist travels 450km in 5 hours. What is his speed in m/s?
First, we convert it to km/h450km / 5h = 90km/hThen, we divide the number by 3.6 to get the answer in m/s90 / 3.6 = 25 m/s-Luca