20 POINTS

A brick is dropped with zero initial speed from the roof of a building and strikes the ground in 1.90 s. How tall is the buildin

irinina [24]

Answer:

17.69 m

Explanation:

The time it takes the brick to strike the ground is 1.90 seconds.

We can apply one of Newton's equation of linear motion to find the height of the building:

$s = ut + 0.5gt^2$

where s = distance (in this case height)

u = initial velocity = 0 m/s

t = time = 1.90 s

g = acceleration due to gravity = 9.8 m/s^2

Therefore:

s = (0 * 1.9) + (0.5 * 9.8 * 1.9 * 1.9)

s = 0 + 17.68

s = 17.69 m

The height of the building is 17.69 m.

6 0

4 years ago

A bike rider was moving to the right at a constant speed. Suddenly the wind then starts blowing against her with 3 N of force to

stiks02 [169]

Answer:

The correct option is D

Explanation:

This question can be better understood when discussed using the Newton's first law of motion which states that an object would continue to move with a uniform speed (in a straight line) unless acted upon by an external force. What happens here (in the question) is that the bike rider would have continued moving at a constant speed (to the right) if not for the opposing force of the wind that acted against her (to the left). <u>This wind/force would cause her speed to reduce or slow down (as posited by the law)</u>.

8 0

3 years ago

A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a

Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ = (3.14-1.97) m = 1.17 m

7 0

3 years ago

Read 2 more answers

FIGURE 1 shows part of a mass spectrometer. The whole arrangement is in a vacuum. Negative ions of mass 2.84 x 10-20 kg and char

yuradex [85]

Yes, the ions can exit slit P without being deflected, if the electric field strength is 170.6 N/C

Explanation:

When the ions are inside the container, they are subjected to two forces, with directions opposite to each other:

The force due to the electric field, whose magnitude is $F_E=qE$ , where q is the charge of the ion and E is the strength of the electric field
The force due to the magnetic field, whose magnitude is $F_B=qvB$ , where v is the speed of the ions and B is the strength of the magnetic field

The ions will move straight and undeflected if the two forces are equal and opposite. By using Fleming Left Hand rule, we notice that the magnetic force on the (negative) ions point upward: this means that the electric field must be also upward (so that the electric force on the ions is downward). Then, the two forces are balanced if

$F_E = F_B$

which translates into

$qE=qvB\\\rightarrow v = \frac{E}{B}$

Therefore, if the speed of the ions is equal to this ratio, the ions will go undeflected.

We can even calculate the value of E at which this occurs. In fact, we know that the ions are earlier accelerated by a potential difference $V=-3000 V$ , so we have that their kinetic energy is given by the change in electric potential energy: