What is the first step in the event of a fire in a building?

A ____________ galaxy has arms of gas and dust; where as _____________ galaxy is a category of galaxy that does not have a disti

Jet001 [13]

Answer:

spiral, an irregular

Explanation:

i took the test :)

3 0

3 years ago

5. If one object has a greater speed than a second object. does the first necessarily have a greater acceleration? Explain, usin

Sveta_85 [38]

Answer:

Explanation:

5. not necessarily so that the first object could have left with initial velocity and the second not, so even if the second has a greater acceleration its velocity is less than that of the first

6. The acceleration of the motorcycle is

SI System Reductions

Vo = 80 km / h (1000m / 1km) (1h / 3600s) = 22.2 m / s

Vf = 90 km / h (1000m / 1km) (1h / 3600s) = 25 m / s

Vf = Vo + at at = Vf-Vo

am = (Vf-Vo) / t

am = (25 -22.2) / t = 2.8 / t

am= 2.8/t

For the bike we have

Vf = 10 km / h (1000m / 1km) (1h / 3600s) = 2.78 m / s

Vo = 0

ab = (Vf -Vo) / t

ab = (2.78 -0) / t

ab = 2.8/t

Since time is the same for both of us, if we round to Significant figures the two accelerations are equal

7. If when an object is slowing or slowing down.

For example, a car goes north and must stop at the traffic light, the acceleration of the brakes goes south

8. Yes, since an object can go to the left and the acceleration to the right, but the object will lose speed over time

9. in the launch of projectiles the acceleration is negative and the speed after half the path is also negative

10. Car B must be moving to car A, because if they leave together B has more acceleration, bone that travels the distance at the same time

11. When we have friction, the velocity of an object increases by an external force, but the friction also increases the acceleration, but since it is positive, the velocity increases until the acceleration is zero and hence the velocity remains constant.

8 0

3 years ago

What is the bouncing of light when it changes media

matrenka [14]

When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.

The part of the light that continues on into the new medium is transmitted
light. Its forward progress at any point in its journey is transmission.

Its direction usually changes as it crosses the boundary. The bending is 
refraction.

The part of the light that bounces away from the boundary and heads back
into the first medium is reflected light. The process of bouncing is reflection.

5 0

2 years ago

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$

6 0

3 years ago

A Christmas sphere will break if dropped on hardwood floor with a velocity of 2.0 m/s. Mar Estefanía is

Margaret [11]

Answer:

0.17 s

Explanation:

Given:

Δy = 0.15 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

0.15 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.17 s

8 0

3 years ago