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3 years ago

A 1,400 kg car accelerates from rest to 30 m/s in 6.0 seconds. what is the net force on the car?

1 answer:

4 0

So your finding acceleration first which is 30m/s divides by 6 seconds equals 5m/s^s and then multiply that by 1,400 kg and you have net force which is 7,000N

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Mathew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the bal

Cloud [144]

When It begins to drop because that when gravity will have its strongest pull on the object.

8 0

3 years ago

What is the total wavelength if one-half of the wave is 3?

seraphim [82]

Answer:

Explanation:

Half the wave = 3

Wavelength = 3 x 2 = 6

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2 years ago

As you stand by the side of the road, a car approaches you at a constant speed, sounding its horn, and you hear a frequency of 8

kap26 [50]

Answer: velocity of the car is 113.33m/s

Explanation:

From Doppler effect,

in the case which the source is moving towards the observer at rest

f2 = v/(v-vs) *f1

where f2 is the final observed frequency

f1 is the initial observed frequency

v = 340m/s (speed of sound in air)

vs = velocity of the source of sound.

rearranging the above equation

f2*(v - vs) = f1* v

vs = (f1* v/f2) - v

but f1 = 80Hz

f2 = 60Hz

v = 340m/s

substituting,

vs = (80 x 340)/60 - 340

vs = 453.33 - 340

vs = 113.33m/s

velocity of the car is 113.33m/s

5 0

3 years ago

The law of reflection states that if the angle of incidence is 32 degrees, the angle of reflection is ___ degrees.

shusha [124]

The angle of reflection is equal to angle of incidence so the angle of reflection is also 32°.

6 0

3 years ago

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

6 0

3 years ago