In what way are all sounds alike?

How do i calculate acceleration? pls help! thanks :)

LenKa [72]

The acceleration is 2 mph/min

Explanation:

The acceleration of an object is the rate of change of velocity. It is defined as follows:

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

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

For the car in this problem, we have (taking North as positive direction):

u = 25 mph

v = 35 mph

t = 5 min

Substituting into the equation, we find the acceleration:

$a=\frac{35 mph-25mph}{5min}=2 mph/min$

Learn more about acceleration:

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#LearnwithBrainly

4 0

3 years ago

For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km

muminat

Answer with Explanation:

We are given that

$y=(18-2x^2) km$

$x=(4t-3)km$

Differentiate x and y w.r.t t

$\frac{dx}{dt}=4$

$\frac{dy}{dt}=-4x\frac{dx}{dt}$

$\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)$

$v_x=\frac{dx}{dt}=4$

$v_y=\frac{dy}{dt}=-16(4t-3)$

Substitute t=1

$v_x=4$

$v_y=-16(4-3)=-16$

Magnitude of velocity= $\mid v\mid=\sqrt{v^2_x+v^2_y}$

$\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s$

Hence, the magnitude of the missile's velocity=16.49 m/s

$a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0$

$a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64$

Substitute t=1

$a_x=0,a_y=-64$

$\mid a\mid=\sqrt{a^2_x+a^2_y}$

$\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2$

Hence, the magnitude of acceleration when t=1 s= $64m/s^2$

3 0

3 years ago

Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert

True [87]

To solve this problem, we know that:

1 Albert = 88 meters

1 A = 88 m

The first thing we have to do is to square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since it is given that 1 acre = 4,050 m^2, so to reach that value, 1st let us divide both sides by 7,744:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then we multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Therefore 1 acre is equivalent to about 0.52 square alberts.</span>

7 0

3 years ago

An investigator places a sample 1.0 cm from a wire carrying a large current; the strength of the magnetic field has a particular

AURORKA [14]

Answer:

she must increase the current by factor of 7

Explanation:

The magnetic field produced by a steady current flowing in a very long straight wire encircles the wire.In order to solve the question, we use this formula,

B= μo I/(2πr)

where,

'μo' represents permeability of free space i.e 4π*10-7 N/A2

B=magnetic field

I= current

r=radius

->When r= 1cm=> 0.01m

B1 = μo $I_{1$ /(2π x 0.01)

->when r=7cm =>0.07m

B2 = μo $I_{2}$ /(2π x 0.07)

Now equating both of the magnetic fields, we have

B1= B2

μo $I_{1$ /(2π x 0.01)= μo $I_{2}$ /(2π x 0.07)

$I_{1$ / $I_{2}$ = 0.01/0.07

$I_{1$ / $I_{2}$ = 1/ 7

Therefore, she must increase the current by factor of 7

5 0

4 years ago

A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.

Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder $(L)=5 m$

Foot the ladder is moving away with speed of $\frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s$

From diagram

$x^2+y^2=L^2$ ------1

at $x=3$

$y^2=25-9=16$

$y=4 m$

Now differentiating equation 1 w.r.t time

$2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}$

$3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s$

negative indicates distance is decreasing with time

5 0

3 years ago