Ask question

Ask question

All categories

3 years ago

14

A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when

the acceleration first began?

1 answer:

Agata [3.3K]3 years ago

6 0

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

$v^2-u^2=2ad$

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

$a=-8.0 m/s^2$ is the acceleration

Solving for u, we find the initial velocity:

$u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s$

You might be interested in

Microwave ovens emit microwave energy with a wavelength of 12.1 cm. What is the energy of exactly one photon of this microwave r

EastWind [94]

Answer:

$1.64\cdot 10^{-24}J$

Explanation:

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelenght of the photon

For the microwave photons in this problem,

$\lambda=12.1 cm=0.121 m$

so their energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.121 m}=1.64\cdot 10^{-24}J$

3 0

2 years ago

The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded togeth

Basile [38]

Answer:

$T=2.5*10^{-3}s$

Explanation:

From the question we are told that:

Beat frequency $F_b=4$

Frequency $F=400Hz$

Generally the equation for Frequency of the violin is mathematically given by

$f_v=F_b+F$

$f_v=4+400Hz$

$f_v=404Hz$

Therefore the period of the violin string oscillations is

$T=\frac{1}{f_v}$

$T=\frac{1}{404}$

$T=2.5*10^{-3}s$

3 0

2 years ago

Please answer of this question

Bogdan [553]

Answer:

$\frac{\pi }{15}$ m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = $2\pi r(\frac{60}{360} )$

= $2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m$

8 0

2 years ago

is the science that manages the database, digital tools, and software used by both medicine and forensic science to store and an

Ne4ueva [31]

Explanation :

Bioinformatics is the science that manages the database, digital tools and software used by both medicine and forensic science to store and analyze DNA.

It involves the use of computers to collect all the data and organize data. It also develops methods and computational tools for understanding biological data.

It helps in tracing the evolution of organism using DNA and builds a computational model.

8 0

3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago