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3 years ago

A rock at rest falls off a tall cliff and hits the valley below after 3.5s. What is the rocks velocity as it hits the ground

Physics

2 answers:

pogonyaev3 years ago

7 0

T = 3.5 secs

Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s

Paraphin [41]3 years ago

5 0

V = u + at
u=0(rest) ; a=gravitational acc(g)=10
v = 0 + 10(3.5) = 35m/s

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

ra1l [238]

Answer:

a) t = 4.16 s

b) x = 141.51 m

Explanation:

Given

v = 21.5 m/s

x0 = 52.0 m

a = 6.0 m/s²

a) Motorcycle

x = v0*t + (a*t²/2)

x = 21.5t + (6*t²/2)

x = 21.5t + 3t² (I)

Car

x = x0 + v0*t

x = 52 + 21.5t (II)

then we can apply I = II

21.5t + 3t² = 52 + 21.5t

⇒ 3t² = 52

⇒ t = 4.16 s

b) We can use I or II, then

x = 52 + 21.5*(4.16)

⇒ x = 141.51 m

8 0

2 years ago

A tourist drops (from rest) a ping pong ball from the top of the tower, which has a height of 324 meters. Assuming no air resist

Elanso [62]

Answer:

8.13secs

Explanation:

From the question weal are given

Height H =324m

Required

time it takes to drop t

Using the equation of motion

H = ut + 1/2gt²

Substitute the given values

324 = 0(t)+1/2(9.8)t²

324 = 1/2(9.8)t²

324 = 4.9t²

t² =324/4.9

t² = 66.12

t = √66.12

t = 8.13secs

Hence the time taken to drop is 8.13secs

4 0

2 years ago

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0

2 years ago

After the body has begun shivering, what explains why it would stop shivering?

Oksana_A [137]

The body shivers to produce energy and it uses the energy to keep it warm. The body would stop shivering when it has produced enough energy to keep it warm and the atmosphere around it has got warmer

3 0

3 years ago

Read 2 more answers

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

4 0

2 years ago