All categories

3 years ago

A car is traveling at 75 m/s. 50 seconds later it is traveling at 25 m/s. What is the car’s acceleration?

Physics

1 answer:

scoray [572]3 years ago

5 0

Answer:

a = -1 m/s^2

Explanation:

Vi = 75 m/s

Vf = 25 m/s

t = 50 s

Plug those values into the following equation:

Vf = Vi + at

25 = 75 + 50a

---> a = -1 m/s^2

You might be interested in

What is the purpose of the scapula to move during arm elevation?

Inessa [10]

The purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

<h3>What is the importance of movement of the scapula during arm elevation?</h3>

The scapula is an important bone which is found in the shoulder and back region of the body.

The scapula enables and increases the range of motion of the arm with its motions.

During arm elevation, the scapula undergoes an upward rotational motion.

Therefore, the purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

Learn more about scapula motion at: brainly.com/question/5133017

#SPJ12

3 0

1 year ago

When a person speaks, a sound intensity is generated that is 600 times greater than when the person whispers. What is the differ

Charra [1.4K]

Answer:

Originally : Level = log I / I0

Currently: Level = 10 log I / I0

Level = 10 log 600 = 10 * 2.78 = 27.8

Note the term 1 bel = 10 decibels

5 0

2 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her pisto

lapo4ka [179]

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift we have

$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}$

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

$\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}$

Solving for $D_{2}$ we get

$D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm$

5 0

3 years ago

Trong thí nghiệm về giao thoa sóng trên mặt nước gồm hai nguồn kết hợp S1S2 cách nhau 15 cm với dao động với tần số 30Hz. Tốc độ

mojhsa [17]

Answer:

this is a difficult question but I will try to answer it answer for this is 3220 a + b b u s y d l new

6 0

2 years ago