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erik [133]
3 years ago
11

If a projectile was launched at an angle of 13 degrees above the ground and hit its target 14.3 m away, what was the projectile'

s launch speed? Ignore air resistance and acceleration due to gravity is 9.8m/s^2.
Physics
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

18 m/s

Explanation:

Range of a projectile on level ground is:

R = v₀² sin(2θ) / g

14.3 m = v₀² sin(2×13°) / 9.8 m/s²

v₀ = 17.9 m/s

Rounded to two significant figures, the launch speed was 18 m/s.

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3 years ago
A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c
Reika [66]
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
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8 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane-parallel plates. An electron is released from
Zigmanuir [339]

Answer:

Explanation:

  • given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
  • q = 1.6 x 10_19C
  • using S = at^2/2
  • acceleration = 0.032 X 2 /(1.30×10−8)^2

a = 3.79 x 10^14m/s^2

  • From F = ma
  • F = qE
  • ma = qE

E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric field. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as

= 2 X 3.79 x 10^14 X 0.032

= 4.92 X 10^6m/s

5 0
2 years ago
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zhannawk [14.2K]
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7 0
3 years ago
Two long straight wires are parallel and 8.6 cm apart. They are to carry equal currents such that the magnetic field at a point
Neko [114]

Answer:

(a) The current should be in opposite direction

(b) The current needed is 39.8 A

Explanation:

Part (a)

Based, on right hand rule, the current should be in opposite direction

Part (b)

given;

strength of magnetic field, B = 370 µT

distance between the two parallel wires, d = 8.6 cm

B = \frac{\mu_oI}{2\pi R}

At the center, the magnetic field strength is twice

B_c = 2(\frac{\mu_oI}{2\pi R}) =\frac{ \mu_oI}{\pi R}

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

B_c = \frac{ \mu_oI}{\pi R}\\\\I = \frac{B_c\pi R}{\mu_o} = \frac{370 *10^{-6}* \pi *0.043}{4\pi *10^{-7}}\\\\I = 39.8 \ A

Therefore, current needed is 39.8 A

6 0
2 years ago
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