What is the speed of a car that traveled 300 miles in 6 hours?

What would be the answer for this and how?

Veronika [31]

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

$k = \frac{F}{\Delta x}\\$

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

$k = \frac{10\ N}{0.04\ m}\\k = 250\ N/m\\$

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

$k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\$

For a load of 30 N, applying Hooke's Law:

$\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\$

Hence, the correct option is:

7 0

2 years ago

A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th

ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

-Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

a is the acceleration of the car

a = $\frac{frictional force}{mass}$ = $\frac{10500}{1500}$ = -7m/s²

Now using;

V² = U² + 2as

V is the final velocity

U is the initial velocity

a is the acceleration

s is the distance moved

0² = 25² + 2 x 7 x s

0 = 625 - 14s

-625 = -14s

s = 44.64m

learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

7 0

3 years ago

1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of

nevsk [136]

Answer:

$4.8 \cdot 10^4 N$

Explanation:

The force of gravity acting on the satellite is given by:

$F=\frac{GMm}{r^2}$

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

$r=2.4\cdot 10^6 m$

Substituting into the equation, we find:

$F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N$

<em>Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).</em>

3 0

3 years ago

Ultraviolet light having a wavelength of 97 nm strikes a metallic surface. Electrons leave the surface with speeds up to 3.48 ×

irga5000 [103]

Answer:

Explanation:

Before calculating the work function, we must know the formula for calculating the kinetic energy of an electron. The kinetic energy of an electron is the taken as the difference between incident photon energy and work function of a metal.

Mathematically, KE = hf - Ф where;

h is the Planck constant

f is the frequency = c/λ

c is the speed of light

λ is the wavelength

Ф is the work function

The formula will become KE = hc/λ - Ф. Making the work function the subject of the formula we have;

Ф = hc/λ - KE

Ф = hc/λ - 1/2mv²

Given parameters

c = 3*10⁸m/s

λ = 97*10⁻⁹m

velocity of the electron v = 3.48*10⁵m/s

h = 6.62607015 × 10⁻³⁴

m is the mass of the electron = 9.10938356 × 10⁻³¹kg

Substituting the given parameters into the formula Ф = hc/λ - 1/2mv²

Ф = 6.63 × 10⁻³⁴*3*10⁸/97*10⁻⁹ - 1/2*9.11*10⁻³¹(3.48*10⁵)²

Ф = 0.205*10⁻¹⁷ - 4.555*10⁻³¹*12.1104*10¹⁰

Ф = 0.205*10⁻¹⁷ - 55.163*10⁻²¹

Ф = 0.205*10⁻¹⁷ - 0.0055.163*10⁻¹⁷

Ф = 0.1995*10⁻¹⁷Joules

Since 1eV = 1.60218*10⁻¹⁹J

x = 0.1995*10⁻¹⁷Joules

cross multiply

x = 0.1995*10⁻¹⁷/1.60218*10⁻¹⁹

x = 0.1245*10²

x = 12.45eV

<em>Hence the work function of the metal in eV is 12.45eV</em>

6 0

3 years ago

How do you calculate distance using deceleration and time?

Effectus [21]

Answer:

U = initial velocity, t = time taken, s = distance covered. Deceleration Formula is used to calculate the deceleration of the given body in motion.

6 0

3 years ago