I did questions 2 and 3 but I don't know if they are right. Someone help!

A group of physics students hypothesize that for an experiment they are performing, the speed of an object sliding down an incli

goblinko [34]

Answer:

v = 2.974

Explanation:

Perhaps the formula should be

v = √(2*g*d (sin(θ) - uk*cos(θ) ) This is a bit easier to read.

v = √(2* 9.80*0.725(0.707 - 0.12*0.707) ) Substitute values. Find 2*g*d

v = √14.21 * (0.707 - 0.0849) Figure out Sin(θ) - uk cos(θ)

v = √14.21 * (0.6222)

v = √8.8422 Take the square root of the value

v = 2.974

6 0

3 years ago

Read 2 more answers

1. A car with a mass of 2500 kg accelerates when the traffic light turns green. If the net force

soldier1979 [14.2K]

;Net force = mass of the body × acceleration of the body due to the net force

; 5000 = 2500 a...then divide both sides by 2500

; acceleration(a) = 2 m/s^2

5 0

3 years ago

A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$

This force work as centripetal force

So $F=\frac{mv^2}{r}$

$4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}$

v = $=0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec$

6 0

3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t

fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

$F = m a\\\\a =\frac{F}{m}$

We use movies and find lips

$\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t$

The moment is defined by

$\to p = m v$

The moment change

$\Delta p = m v - m v_0$

Let's replace the speeds in this equation

$\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t$

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0

3 years ago

A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0

nikklg [1K]

Answer:

(a) 10 s

(b) 30 m/s

(c) 150 m

Explanation:

The motorist's position at time t is:

x = 15t

The officer's position at time t is:

x = ½ (3) t² = 1.5 t²

(a) When they have the same position, the time is:

15t = 1.5 t²

t = 0 or 10 s

(b) The officer's speed is:

v = 3t

v = 30 m/s

(c) The position is:

x = 15t = 150 m

6 0

3 years ago