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Marizza181 [45]
3 years ago
12

A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

ullet hits the ground?(b) How far does the bullet travel horizontally?
Physics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a)  Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Displacement, s = 1.5 m      

     Substituting

                      s = ut + 0.5 at²

                      1.5 = 0 x t + 0.5 x 9.81 xt²

                      t = 0.553 s

      Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 200 m/s

        Acceleration, a = 0 m/s²  

        Time, t = 0.553 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 200 x 0.553 + 0.5 x 0 x 0.553²

                      s = 110.6 m

      The bullet travels horizontally 110.6 m

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Explanation:

Let m be the mass of a little car and m' be the mass of another car.

We know that,

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ATQ,

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a = 2 × a'

a'=\dfrac{a}{2}\\\\a'=\dfrac{2.57}{2}\\\\a'=1.285\ m/s^2

So, the acceleration of another little car is equal to 1.285\ m/s^2.

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3 years ago
How do magnet effect materials that are not magnets
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Answer: Magnetic metals & non-magnetic metals both play an important role in engineering. Magnetism is the basis for many applications. At the same time, this property may also be unwanted in certain circumstances.

Therefore, it is important to know which metals are magnetic and which ones are not.

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3 0
3 years ago
The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current
Komok [63]

Answer:

I=0.047A

Explanation:

Let's use Ohm's law:

V=IR  

or

I=\frac{V}{R}   (1)

Where:

V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

R=\rho*\frac{l}{A}    (2)

Where:

R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}

Keep in mind that the electrical resistivity of the gold is a known constant which is \rho_g_o_l_d=2.35*10^{-8} and the cross sectional area of the conductor is calculated as:

A=\pi *(r^{2})=\pi  *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} }  =14.96056465\Omega

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

I=\frac{0.7}{14.96056465}\approx0.047A

7 0
3 years ago
A coaxial cable consists of an inner conductor with radius ri = 0.20 cm and an outer radius of ro = 0.4 cm and has a length of 1
N76 [4]

Answer:

R = 1.69*10^{11} ohm

Explanation:

ri = 0.20cm

ro = 0.4 cm

length L = 13m

resistivity \rho = 2.00*10^13 ohm m

resistance can be determine by using following relation

R = \frac{\rho}{2\pi L} ln\frac{ro}{ri}

R = \frac{2.00*10^{13}}{2\pi *10} ln\frac{.004}{.002}

R = 1.69*10^{11} ohm

4 0
3 years ago
Una tractomula se desplaza con rapidez de 69 km/h. Cuando el conductor ve una vaca atravesada enmedio de la carretera, acciona l
Anestetic [448]

Answer:

Los datos que tenemos:

Rapidez: 69km/h

Tiempo que tarda en frenar = 4s.

Distancia inicial entre la tracto-mula y la vaca = 25m

Ok, la ecuación de desaceleración es:

D = (sf - si)/t

sf = velocidad final = 0m/s

si = velocidad inicial = 69km/h

t = tiempo = 4s

D = -69km/h/4s

ok, 1h = 3600s

D = (-69km/s)*1/(4*3600s)  = -0.0048 km/s^2

Entonces la ecuación de aceleración es:

a(t) =  -0.0048 km/s^2

Para la velocidad, integramos sobre el tiempo

v(t) = (-0.0048 km/s^2)*t + v0

donde v0 es la velocidad inicial, en este caso v0 = 69km/3600s = 0.0191km/s  

v(t) =  (-0.0048 km/s^2)*t + 0.0191km/s

Para la posición volvemos a integrar sobre el tiempo, esta vez suponemos la posición inicial igual a cero.

p(t) = (1/2)*(-0.0048 km/s^2)*t^2 + 0.0191m/s*t

Ahora, si p(t=4s) < 25m, esto implica que la tracto-mula no impacto con la vaca.

p(4s) = (1/2)*(-0.0048 km/s^2)*(4s)^2 + 0.0191km/s*4s = 0.038km

y 1km = 1000m

0.038km = 0.038*1000m = 38m

Entonces si, atropello a la vaca.

4 0
4 years ago
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