All categories

3 years ago

B. You are approaching a stoplight moving with a velocity of 42.0 m/s. The light turns

Physics

2 answers:

ddd [48]3 years ago

8 0

Answer:

Time needed is 2.5 seconds. The distance covered is 31.3 m

Explanation:

Finger [1]3 years ago

3 0

Answer:

displacement of the car will be 88.2 m

Explanation:

As we know that initial speed of the car is given as

$v_i = 42 m/s$

after complete stop the final speed of the car is

$v_f = 0$

now the deceleration caused due to applied brakes is given as

$a = 10 m/s^2$

now by kinematics we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 42^2 = 2(-10)d$

$d = 88.2 m$

You might be interested in

Luigi twirls a round piece of pizza dough overhead with a frequency of

viva [34]

The linear speed of the pepperoni is 0.628 m/s. Its direction is tangential to the circle.

We know that;

v = rω

r = radius of the piece = 10 cm or 0.1 m

ω = angular velocity

We have to convert 60 revolutions per minute to radians per second

1 rev/min = 0.10472 rad/s

60 revolutions per minute = 60 rev/min × 0.10472 rad/s/1 rev/min

= 6.28 rad/s

v = 0.1 m × 6.28 rad/s

v = 0.628 m/s

The direction of this velocity is tangential to the circle.

Learn more: brainly.com/question/4612545

5 0

2 years ago

Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon

Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0

3 years ago

A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exe

PtichkaEL [24]

Answer:

The pressure exerted by the brick on the table is 18,933.3 N/m².

Explanation:

Given;

height of the brick, h = 0.1 m

density of the brick, ρ = 19,300 kg/m³

acceleration due to gravity, g = 9.81 m/s²

The pressure exerted by the brick on the table is calculated as;

P = ρgh

P = (19,300)(9.81)(0.1)

P = 18,933.3 N/m²

Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².

4 0

3 years ago

An electron in the first energy level of the electron cloud has an electron in the third energy level

lukranit [14]

Answer:

a lower energy than

Explanation:

3 0

3 years ago

Read 2 more answers

What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e

Zinaida [17]

Answer:

$F=1.38*10^{-9}N$

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

Here k is the Coulomb constant. In this case, we have $q_1=-e$ , $q_2=e$ and $d=4.09*10^-10m$ . Replacing the values:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N$

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

$F=1.38*10^{-9}N$

5 0

3 years ago