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diamong [38]
2 years ago
8

A missile is launched vertically from a missile silo, it will explode after 32 s. It's launch speed was 145 m/s, and it doesn't

accelerate after launch. Will the missile have hit its highest point before it explodes? Will it have landed before it explodes?​
Physics
1 answer:
Karolina [17]2 years ago
4 0

Answer:

Landed before it explodes

Explanation:

vf = vi + at,

0 = 145 - (9.8)t,

t = 14.79 s (Time to reach highest point)

14.79 x 2 = 29.59 s (Time to land on the ground)

It will have landed before it explodes because both the time to reach the highest point and the time to land on the ground are less than 32 seconds.

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A crash test dummy is inside a test car moving at 55 km/h. How fast is the dummy moving relative to the seat he is sitting in?
kvasek [131]

The dummy is moving with a speed 0 km/h relative to the seat in which it is sitting.

If the relative speed was non-zero, the dummy would move away from its seat, which contradicts the problem formulation.

7 0
2 years ago
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How large a band of frequencies does each television broadcasting channel get ?
k0ka [10]

Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

8 0
2 years ago
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1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
2 years ago
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal
AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}

vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0
3 years ago
A bar on a hinge starts from rest and rotates with an angular acceleration α = 12 + 5t, where α is in rad/s2 and t is in seconds
Rama09 [41]

Answer:

Ф = 239.73 rad

Explanation:

α = 12 + 15×t

W = ∫α×dt

   = ∫(12 + 5×t)×dt

   = 12×t + 2.5×t^2

then:

Ф = ∫W×dt

   = ∫(12×t + 2.5×t^2)dt

   = 6×t^2 + 5/6×t^3

therefore the angle at t = 4.88s is:

Ф = 6×(4.88)^2 + 5/6×(4.88)^3

   = 239.73 rad    

5 0
2 years ago
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