A toy car has a mass of 1500g. If a 3-N force is applied to the car, what will it's acceleration be?

A 10,000 N net force is accelerating a car at a rate of 5.5m/s^2. What is the cars mass

ycow [4]

Answer:

If a net horizontal force of 175 N is applied to a bike whose mass is 43 kg what acceleration is produced? What average net force is ... A 10,000 N net force is accelerating a car at a rate of 5.5 m/s2. What is the car's mass? A boy pedals his ...

Explanation:

6 0

3 years ago

Which statement explains the similarity between momentum and kinetic energy of an object?

Anon25 [30]

Answer:B.

Both increase as the mass and velocity increase.

3 0

3 years ago

Read 2 more answers

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

3 years ago

A ray of light passes from one material into a material with a higher index of refraction. Determine whether each of the followi

nikitadnepr [17]

Answer:

a) the Angle te also decreases , b) decrease, c) unchanged , d) the speed decrease , e) unchanged

Explanation:

When a ray of light passes from one transparent material to another, it must comply with the law of refraction

n1 sin θ₁ = n2 sin θ₂

Where index v1 is for the incident ray and index 2 for the refracted ray

With this expression let's examine the questions

a) They indicate that the refractive index increases,

sin θ₂ = n₁ / n₂ sin θ₁

θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)

As m is greater than n1 the quantity on the right is less than one, the whole quantity in parentheses decreases so the Angle te also decreases

Answer is decrease

b) The wave velocity eta related to the wavelength and frequency

v = λ f

The frequency does not change since the passage from one medium to the other is a process of forced oscillation, resonance whereby the frequency in the two mediums is the same.

The speed decreases with the indicated refraction increases and therefore the wavelength decreases

λ = λ₀ / n

The answer is decrease

c) from the previous analysis the frequency remains unchanged

d) the refractive index is defined by

n = c / v

So if n increases, the speed must decrease

The answer is decrease

e) the energy of the photon is given by the Planck equation

E = h f

Since the frequency does not change, the energy does not change either

Answer remains unchanged

7 0

3 years ago

Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre

castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619

=

t

+

C

Here C =constant of integration.

x

=

0 at t

=

0

, we get: C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619 =

t

−

4.7619

.

(

2

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= −

4.7619

=

1

−

4.7619

= −

3.7619

or x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in m

i

l

/

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

v

=

7.5×

Now differentiating above equation w.r.t time we get:

a

=

d

v/

d

t

=

−

0.675

/

At

t

=

0

the joggers acceleration is :

a

=

−

0.675

m

i

l

/

=

−

4.34

×

f

t

/

(c) required time for the jogger to run 6 miles is obtained by setting

x

=

6 in equation (2). We get:

−

4.7619

(

1

−

(

0.04

×

6 )

)^

7

/

10=

t

−

4.7619

or

t

=

0.832

h

r

s

6 0

3 years ago