Ask question

Ask question

All categories

3 years ago

7

The muzzle velocity of an armor-piercing round fired from an M1A1 tank is 1770 m/s (nearly 4000 mph or mach 5.2). A tank is at t

he top of a cliff and fires a shell horizontally. If the shell lands 6520 m from the base of the cliff, how high is the cliff?

1 answer:

loris [4]3 years ago

8 0

<h2>Height of cliff is 66.43 meter.</h2>

Explanation:

Consider the horizontal motion of shell,

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 1770 m/s

Acceleration, a = 0 m/s²

Displacement, s = 6520 m

Substituting

s = ut + 0.5 at²

6520 = 1770 x t + 0.5 x 0 xt²

t = 3.68 s

Time of flight of shell = 3.68 s.

Now consider the vertical motion of shell,

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3.68 s

Substituting

s = ut + 0.5 at²

s = 0 x 3.68 + 0.5 x 9.81 x 3.68²

s = 66.43 m

Height of cliff is 66.43 meter.

You might be interested in

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

3 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

3 years ago

If you wish to observe features that are around the size of atoms, say 1 .5 x 100 m, with electromagnetic radiation, the radiati

chubhunter [2.5K]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

If you wish to observe features that around the size of atoms, say 1.5×10⁻¹⁰ m, with electromagnetic radiation, the radiation must have a wavelength about the size of the atom itself.

a) If you had a microscope which was capable of doing this, what would the frequency of electromagnetic radiation be, in hertz that you would have to use?

b) What type of electromagnetic radiation would this be?

Given Information:

Wavelength = λ = 1.5×10⁻¹⁰ m

Required Information:

a) Frequency = f = ?

b) Type of electromagnetic radiation = ?

Answer:

a) Frequency = f = 2×10¹⁸ Hz

b) Type of electromagnetic radiation = X-rays

Explanation:

a) The frequency of the electromagnetic radiation is given by

f = c/ λ

Where λ is the wavelength of the electromagnetic radiation and c is the speed of light and its value is 3×10⁸ m/s

f = 3×10⁸/1.5×10⁻¹⁰

f = 2×10¹⁸ Hz

Therefore, the frequency of the electromagnetic radiation would be 2×10¹⁸ Hz.

b)

The frequency range of X-rays is 3×10¹⁶ Hz to 3×10¹⁹ Hz

The frequency 2×10¹⁸ lies in that range, therefore, the type of electromagnetic radiation is X-rays

5 0

3 years ago

What is the value of x in the equation below 1+2e^x+1=9

GuDViN [60]

<h2>Answer: 1.252</h2>

Explanation:

We are given this equation and we need to find the value of $x$ :

$1+2e^x+1=9$ (1)

Firstly, we have to clear $x$ :

$2e^x=9-1-1$

$2e^x=7$

$e^x=\frac{7}{2}$ (2)

Applying<u> Natural Logarithm</u> on both sides of the equation (2):

$ln(e^x)=ln(\frac{7}{2})$ (3)

$xln(e)=ln(\frac{7}{2})$ (4)

According to the Natural Logarithm rules $xln(e)=x$ , so (4) can be written as:

$x=ln(\frac{7}{2})$ (5)

Finally:

$x=1.252$

3 0

3 years ago

Sam moves a box with a with a force of 400N a distance of 5 meters. How long did it take him to move the box if 20 Watts of powe

Vlad1618 [11]

Answer:

100s

Explanation:

there are many student how can not get answer on time and step by step. so there are a group of trusted physics experts who provide step by step answer. just join this wats up group.

4 0

3 years ago

Read 2 more answers