VERY EASY QUESTION FOR HIGH SCHOOL STUDENTS:

A car is initially moving at 10.5 m/s and accelerates uniformly to reach a speed of 21.7 m/s in 4.34 s. How far did the car move

Zarrin [17]

Answer:

A. 69.9m

Explanation:

Given parameters:

Initial velocity = 10.5m/s

Final velocity = 21.7m/s

Time = 4.34s

Unknown:

Distance traveled = ?

Solution:

Let us first find the acceleration of the car;

Acceleration = $\frac{v - u}{t}$

v is final velocity

u is initial velocity

t is the time

Acceleration = $\frac{21.7 - 10.5}{4.34}$ = 2.58m/s²

Distance traveled;

V² = U² + 2aS

21.7² = 10.5² + 2 x 2.58 x S

360.64 = 2 x 2.58 x S

S = 69.9m

3 0

3 years ago

Analyzing a blood sample usually involves which type of separation method?

schepotkina [342]

Answer:(D.)

Explanation:

4 0

2 years ago

A quantity that has direction as well as size is a scalar.<br><br><br> True or False

lisov135 [29]

Answer:

A quantity that does not depend on the direction is called a scalar quantity. Vector quantities have two characteristics, a magnitude, and a direction. Scalar quantities have only a magnitude. When comparing two vector quantities of the same type, you have to compare both the magnitude and the direction.

Scalar quantities only have magnitude (size). Scalar quantities include distance...

A quantity that is specified by both size and direction is a vector. Displacement includes both size and direction and is an example of a vector. However, distance is a physical quantity that does not include a direction and isn't a vector.

Explanation:

hope this helps...

3 0

2 years ago

Read 2 more answers

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

4 0

3 years ago

Imagine that asteroid A that has an escape velocity of 50 m/s. If asteroid B has twice the mass and twice the radius, it would h

padilas [110]

Answer:

The same as the escape velocity of asteorid A (50m/s)

Explanation:

The escape velocity is described as follows:

$v=\sqrt{\frac{2GM}{R}}$

where $G$ is the universal gravitational constant, $M$ is the mass of the asteroid and $R$ is the radius

and since the scape velocity is 50m/s:

$50m/s=\sqrt{\frac{2GM}{R}}$

Now, if the astroid B has twice mass and twice the radius, we have that tha mass is: $2M$

and the radius is: $2R$

inserting these values into the formula for escape velocity:

$v=\sqrt{\frac{2G(2M)}{2R} } =\sqrt{\frac{4GM}{2R} } =\sqrt{\frac{2GM}{R} }$

and we have found that $50m/s=\sqrt{\frac{2GM}{R}}$ , so the two asteroids have the same escape velocity.

We found that the expression for escape velocity remains the same as for asteroid A, this because both quantities (radius and mass) doubled, so it does not affect the equation.

The answer is

Asteroid B would have an escape velocity the same as the escape velocity of asteroid A

7 0

3 years ago