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3 years ago

6. A Cadillac Escalade has a mass of 2 569.6 kg, if it accelerates at 4.65m/S force on the car?

Physics

2 answers:

SVETLANKA909090 [29]3 years ago

5 0

Answer:

F= 2569.6 X 4.65 = 11,948.64

*Multiply the mass and the acceleration to find the force

Explanation:

Roman55 [17]3 years ago

3 0

Answer:11948.64kgm/s^2

Explanation:

Force=mass x acceleration

Force=2569.6 x 4.65

Force=11948.64kgm/s^2

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2 years ago

Read 2 more answers

A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y

Genrish500 [490]

Answer:

Positive z direction.

Explanation:

The magnetic force acting on the electron is given by the formula as :

$F=q(v\times B)$

q is the charge on proton

v is the speed of proton

B is the magnetic field

It is mentioned that the proton is moving with a velocity in the positive x-direction. The uniform magnetic field B in the positive y-direction such taht,

q = +e

v = vi

B = Bj

$F=e(v\ i\times B\ j)$

Since, $i\times j=k$

$F=(evB)\ k$

So, the magnetic force acting on the proton in positive z axis. Hence, the correct option is (d) "positive z direction".

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3 years ago

Monochromatic light of 605 nm falls on a single slit, which is located 85 cm from a

Anna [14]

Answer:

mhm

Explanation:

The answer is c 00.095

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3 years ago

A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.

tankabanditka [31]

Answer:

$v_o=40.14\ m/s$

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

The horizontal speed is constant and equal to the initial speed $v_o$
The vertical speed is zero at launch time, but increases as the object starts to fall
The height of the object gradually decreases until it hits the ground
The horizontal distance where the object lands is called the range

We have the following formulas

$\displaystyle v_x=v_o$

$\displaystyle x=v_o.t$

$\displaystyle v_y=g.t$

$\displaystyle y=\frac{gt^2}{2}$

Where $v_o$ is the initial horizontal speed, $v_y$ is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

$\displaystyle y=\frac{gt^2}{2}$

Rearranging and solving for t

$\displaystyle 2y=gt^2$

$\displaystyle t^2=\frac{2\ y}{g}$

$\displaystyle t=\sqrt{\frac{2\ y}{g}}$

We then replace this value in

$\displaystyle x=v_o.t$

To get

$\displaystyle v_o=\frac{x}{t}$

$\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}$

$\displaystyle v_o=\sqrt{\frac{g}{2y}}.x$

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing $v_o$

$\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6$

The launch velocity is

$\boxed{v_o=40.14\ m/s}$

7 0

3 years ago

A 50.0-kg crate is being pulled along a horizontal smooth surface. The pulling force is 10.0 n and is directed 20.0° above the h

steposvetlana [31]

Answer:

$0.188 m/s^2$

Explanation:

Assuming the crate does not lift above the ground and remains along the floor, then its acceleration will be in the horizontal direction. Therefore, we can use Newton's second law to find its acceleration:

$F_x = ma_x$