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Dima020 [189]
2 years ago
12

Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic​

Physics
1 answer:
Alinara [238K]2 years ago
7 0

The volume of the room is the product of its dimensions:

10\times 12 \times 4 = 480\text{ m}^3

Now, from the equation

d=\dfrac{m}{V}

where d is the density, m is the mass and V is the volume, we deduce

m=dV

So, multiply the density and the volume to get the mass of air in the room.

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A marble is dropped straight down from a distance h above the floor.
Mrac [35]
Fm=Fe and am>ae
Hopefully this helps
7 0
2 years ago
As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -
Svetllana [295]

Answer:

The amount of charge the space shuttle collects is -1.224nC

Explanation:

The magnitude of Electric potential is given as;

V = kq/r

where;

V is the electric potential in volts

k is coulomb's constant

r is the radius of the sphere or distance moved by the charge

given;  V = -1.1 V, k = 8.99 x 10⁹ Nm²/C², r = 10m

Substituting this values in the above equation, we estimate the amount of charge space shuttle collects.

q = (V*r)/k

q = (-1.1 *10)/(8.99 x 10⁹ )

q = -1.224 X 10⁻⁹ C

q = -1.224nC

Therefore, the amount of charge the space shuttle collects is -1.224nC

8 0
2 years ago
A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that
gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

v(t)=At+Bt^2

Differentiating with respect to time

\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt

At t = 0

1.4=A

Hence, A = 1.4 m/s²

B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3

B = -0.10493 m/s³

At t = 5 seconds

a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2

a = 1.29507 m/s²

T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N

T = 28095.8271 N

Weight of rocket

W=2530\times 9.81=24819.9\ N

\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W

T = 1.13198 W

3 0
2 years ago
A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas
goldfiish [28.3K]
    <span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400

This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.

Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.

Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600

This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.

Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)

The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.

x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
3 0
2 years ago
In this lab, you will use a dynamics track to generate collisions between two carts. If momentum is conserved, what variable cha
BartSMP [9]

In collision type of problems since momentum is always conserved

we can say

m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}

So here along with this equation we also required one more equation for the restitution coefficient

v_{2f} - v_{1f} = e(v_{1i} - v_{2i})

so above two equations are required to find the velocity after collision

here the change in velocity occurs due to the contact force while they contact in each other

so this is the impulse of collision while they are in contact with each other while in collision which changes the velocity of two colliding objects

8 0
2 years ago
Read 2 more answers
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