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2 years ago

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A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet to strike a target 3240 meters away

?

2 answers:

Alik [6]2 years ago

7 0

4.5 seconds you devoured 3240/720=4.5

Nataliya [291]2 years ago

4 0

Answer:

Time taken, t = 4.5 seconds

Explanation:

Speed of the bullet, v = 720 m/s

Distance to be covered by the target, d = 3240 m

Let t is the time required for the bullet to strike a target. The time taken is calculated as :

$t=\dfrac{d}{v}$

$t=\dfrac{3240\ m}{720\ m/s}$

t = 4.5 seconds

So, the time required for the bullet to strike a target is 4.5 seconds. Hence, this is the required solution.

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jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

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2 years ago

In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad

Crank

Answer:

a) Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c) Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

We replace

ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

Q = A v

Q = A₁ v₁

Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

A₁ = π r₁

A₂ = π r₂

We replace

Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

r₁ = d₁ / 2 = 1.00 / 2

r₁ = 0.500 10⁻² m

r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

Q = 1.57 10⁻² √(12 10³/2550)

Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

Q = 1.57 10⁻² √(2 12 10³ / (850 3)

Q = 4.8 10⁻² m³ / s

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2 years ago

Cassy shoots a large marble (Marble A, mass: 0.06 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marbl

Lostsunrise [7]

Conservation of linear momentum:

m*v inital = m*v final

0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v

(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)

0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s

Answer 1.8 m/s (positive, to the right).

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2 years ago

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) $1\times10^6N/C$

B) $2\times10^{-13}N$

C) $1\times10^{-2}V$

D) Toward the inner wall

E) $3\times10^{-17}J$

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula $E=\frac{\sigma}{k\epsilon_0}$ , where $\epsilon_0=8.85\times10^{-12}C/Vm$ and in our case $\sigma=10^5C/m^2$ and, for air, $k=1.00059$ , so we have:

$E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C$

B) The K+ ion has one elemental charge excess, so its charge is $q=1.6\times10^{-19}C$ , and the force a charge experiments under an electric field E is given by F=qE, so we have:

$F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N$

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by $\Delta V=dE$ , so we have:

$\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V$

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference $\Delta V$ on a charge Q is $W=Q\Delta V$ , and is equal to the kinetic energy imparted on it, so we have:

$K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J$

5 0

2 years ago

1. Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (m²), what's

Anit [1.1K]

Answers:

1. Firstly, we have to define that Pressure $P$ is Force applied $F$ per unit area $A$ . It is mathematically expressed as follows:

$P=\frac{F}{A}$ (1)

The unit of P is Pascal (Pa) which is equivalent to $\frac{kg}{ms^{2}}$ and also equivalent to $\frac{N}{m^{2} }$

There is also another expression of the Pressure in which it is dependent on the density $d$ of the liquid, the height $h$ of the container and the gravity force $g$ :

$P=d*h*g$ (2)

In this problem the liquid is water, and its known density is approximately:

$d=1000kg/m^{3}$

So, we have to substitute the values in equation (2) to obtain the pressure <u>(Being careful with the units)</u>:

$P=1000\frac{kg}{m^{3}}*10m*9.8\frac{m}{s^{2}}$

$P=98000Pa$

Then, we have to substitute this value in equation (1) and clear $F$ :

$F=P*A$

Finally:

$F=196000N$

2. For this problem, we will use equation (1) to find the Pressure. We already know the area $A$ and the force exerted by water in the container $F$ :

$P=\frac{F}{A}=\frac{900N}{3m^{2}}$

$P=300Pa$

3. In this case, equation (2) is the perfect way to find the hydrostatic pressure at any point at the bottom of the tank <u>(be careful with the units):</u>

$P=d*h*g$

$P=1000\frac{kg}{m^{3}}*7.5m*9.8\frac{m}{s^{2}}$

$P=73500Pa$

4. In this case, it's important to know that in fluids (in this case the water) the higher the fluid is, the lower the pressure. Then, if $P_{1}$ and $P_{2}$ are the respective pressures at the heights $h_{1}$ and $h_{2}$ , and knowing that the water density and the gravity force in this case are constants, we can use the following expression to solve this problem:

$P_{2}- P_{1} =d*g(h_{2}- h_{1})$ (3)

Where:

$P_{1}=1.5 kPa$ at $h_{1}=2m$

Note that $1kPa=1*1000 Pa$

And $P_{2}=?$ is unknown at a given height $h_{2}=6m$

Then, we have to substitute the values in equation (3) to find $P_{2}$ :

$P_{2}-1500Pa=1000\frac{kg}{m^{3}}*9.8\frac{m}{s^{2}} (6m-2m)$

Finally: $P_{2} =40700Pa$

5. In this case we have the area $A=0.75m^{2}$ and the mass of the piston $m=200kg$ , and we need to know the pressure $P$ .

We will use equation (1):

$P=\frac{F}{A}$

But, <u>do you remember that above we stated that pressure is the force applied over an area?</u>

Well, in this case we will use the following equation, in which the gravity force and the mass of a body are involved, to find $F$ :

$F=m*g=200kg*9.8\frac{m}{s^{2}}$

Then:

$F=1960N$

Now we can finally calculate $P$ :

$P=\frac{1960N}{0.75m^{2}}$

$P=2613.33Pa$

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2 years ago

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