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3 years ago

Can someone please help, ;w;?

Physics

2 answers:

SOVA2 [1]3 years ago

7 0

I believe the answer is 3). The cell wall provides protection, it doesn’t control movements of materials in and out of the cell.

damaskus [11]3 years ago

5 0

The correct answer is 3

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In the early twentieth century, _______ redefined psychology as "the science of observable behavior."

Paha777 [63]

Explanation:

Answer :

A. John B. Watson

4 0

3 years ago

A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocit

Dmitry [639]

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

6 0

2 years ago

The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi

SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

$\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3$

$\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3$

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0

3 years ago

The diagram below shows the velocity vectors for two cars that are moving relative to each other.

scoundrel [369]

Answer:

The answer is " $5 \ \frac{m}{s} \ west$ "

Explanation:

$\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\$

velocity of car | respect to car :

$\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\$

$=\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west$

7 0

3 years ago

The apparent height of a building 10.5 km away is 0.02 radians. What is the approximate height of the building to the nearest me

Ksenya-84 [330]

Answer:

Approximate height of the building is 23213 meters.

Explanation:

Let the height of the building be represented by h.

0.02 radians = 0.02 × $\frac{180^{o} }{\pi }$

= 0.02 x (180/ $\frac{22}{7}$ )

0.02 radians = 1.146°

10.5 km = 10500 m

Applying the trigonometric function, we have;

Tan θ = $\frac{opposite}{adjacent}$

So that,

Tan 1.146° = $\frac{h}{10500}$

⇒ h = Tan 1.146° x 10500

= 2.21074 x 10500

= 23212.77

h = 23213 m

The approximate height of the building is 23213 m.

8 0

3 years ago