Find the force needed to accelerate a .3 kg bullet at 2100 m / s / s.

Please help me with this (with explanation)

Sergeeva-Olga [200]

Suppose the cyclist travels for a total time of t hours.

For 20 min = 1/3 hr, the cyclist does not move.

Over the remaining (t - 1/3) hr, the cyclist is moving at a constant speed of 22.0 km/hr, so that the cyclist would travel a distance of

x = (22.0 km/hr) • ((t - 1/3) hr) ≈ (22.0 km/hr) t - 7.33 km

If the cyclist's average speed over the total time t was 17.5 km/hr, then by the definition of average speed,

17.5 km/hr = x / t

Replace x with the distance expression from earlier:

17.5 km/hr = ((22.0 km/hr) t - 7.33 km) / t

Solve for t :

17.5 km/hr = 22.0 km/hr - (7.33 km) / t

(7.33 km) / t = 4.5 km/hr

t = (7.33 km) / (4.5 km/hr)

t ≈ 1.62963 hr

Then the distance the cyclist traveled must have been

x ≈ (22.0 km/hr) (1.62963 hr) - 7.33 km ≈ 28.5 km

and so the answer is A.

Alternatively, as soon as you arrive at

17.5 km/hr = x / t

you can instead solve for t in terms of x, then plug that into the distance equation.

t = x / (17.5 km/hr)

then

x ≈ (22.0 km/hr) (x / (17.5 km/hr)) - 7.33 km

x ≈ 1.25714 x - 7.33 km

0.25714x ≈ 7.33 km

x = (7.33 km) / 0.25714 ≈ 28.5 km

6 0

3 years ago

An object moving north with an initial velocity of 14 m/s accelerates 5 m/s2 for 20 seconds. What is the final velocity of the o

slavikrds [6]

Answer:

Final velocity of the object(v) = 114 m/s

Explanation:

Given:

Initial velocity (u) = 14 m/s

Accelerates (a) = 5 m/s²

Time taken = 20 seconds

Find:

Final velocity of the object(v) = ?

Computation:

According to 1st law of motion.

⇒ v = u +at

⇒ v = 14 + (5)(20)

⇒ v = 14 + 100

⇒ v = 114 m/s

Final velocity of the object(v) = 114 m/s

8 0

3 years ago

__use coherent light.

miv72 [106K]

I need Explanation please

5 0

4 years ago

A charge q of 1.3 × 10-16 coulombs moves from point A to a lower potential at point B in an electric field of 3.2 × 102 newtons/

LUCKY_DIMON [66]

Explanation :

It is given that,

Charge, $q=1.3\times 10^{-16}\ C$

Electric field, $E=3.2\times 10^2\ N/C$

Distance, $d=1.1\times 10^{-2}\ m$

The work done is stored in the form of potential energy.

$W=F.d$

$\because F=qE$

So, $W=qE\ d$

$W=1.3\times 10^{-16}\ C\times 3.2\times 10^2\ N/C\times 1.1\times 10^{-2}\ m$

$W=4.576\times 10^{-16}\ J$

Hence, this is the required solution.

5 0

3 years ago

Read 2 more answers

Cassy shoots a large marble (Marble A, mass: 0.06kg) at a smaller marble (Marble B, mass: 0.03kg) that is sitting still. Marble

4vir4ik [10]

The question can be solved using conservation of linear momentum.

$M_{a}$ = 0.06kg and $M_{b}$ = 0.03kg

Let the initial velocity of Marble A be , $V_{a1}$ = 0.7m/s

Let the initial velocity of Marble B be, $V_{b1}$ = 0m/s

Let the velocity of Marble A after collisiong , $V_{a2}$ = -0,02m/s

Let the velocity of Marble B after collision be $V_{b2}$

From the conservation of linear momentum equation. We get,

$M_{a}V_{a1}+M_{b}V_{b1}=M_{a}V_{a2}+M_{b}V_{b2}$

Substituting the values we get,

(0.06)(0.7) + 0 = (0.06)(-0.02) + (0.03) $V_{b2}$

we get, $V_{b2}$ = 1.44m/s

6 0

3 years ago