All categories

3 years ago

A car is traveling at a velocity of 7.5 m/s. How far will it get in 34 sec

Physics

1 answer:

Stella [2.4K]3 years ago

6 0

Answer:

255 m

Explanation:

The car is moving by uniform motion (constant velocity), so we can calculate the distance it travels by using the equation

$d=vt$

where

d is the distance travelled

v is the velocity of the car

t is the time interval

In this problem:

v = 7.5 m/s is the velocity of the car

t = 34 s is the time interval

Therefore, the distance covered by the car is

$d=(7.5)(34)=255 m$

You might be interested in

A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude

Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$ .

Now, let us call $\lambda$ the charge per unit length, then we know that

$dQ = \lambda Rd\theta$ ;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda }{R}.$

Now, we know that

$\lambda = 3.0*10^{-9}C/m,$

$k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},$

and the radius of the semicircle is

$\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };$

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9}) }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0

2 years ago

What is a asteroid traveling rapidly called

SCORPION-xisa [38]

Answer:

meteor

Explanation:

A asteroid stays still and a meteor goes fast

4 0

2 years ago

Read 2 more answers

Non examples of gravity ?

Afina-wow [57]

Floating. When you have no gravity you have nothing to be pushing you down to the floor so that would be an example of no gravity pushing on you.

3 0

2 years ago

Read 2 more answers

A simple circuit consists of a light bulb connected to the terminals of a battery. A voltmeter shows a potential difference of 1

larisa [96]

<h2>Answer:</h2>

38.14Ω

<h2>Explanation:</h2>

Let's solve this question using Ohm's law which states that the current (I) flowing through a conductor is directly proportional to the potential difference or voltage (V) across it. Mathematically;

V = I R -------------------(i)

<em>Where</em>;

R is the constant of proportionality called resistance of the conductor and is measured in Ohms (Ω)

<em>From the question;</em>

V = 18.5V

I = 0.485A

<em>Substitute these values into equation (i) as follows;</em>

18.5 = 0.485 x R

<em>Solve for R;</em>

R = 18.5 / 0.485

R = 38.14Ω

Therefore the resistance of the bulb is 38.14Ω

6 0

2 years ago

A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0

3 years ago