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2 years ago

Speed of 26.7 m/s in 3.06 s. How far had the car traveled by the time the final speed was achieved?

Physics

1 answer:

Alik [6]2 years ago

6 0

Answer:

Below

Explanation:

You can use this equation to find the distance :

distance = velocity x time

distance = (26.7)(3.06)

= 81.702 m

Rounding to 3 sig figs

= 81.7 m

Hope this helps

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HELP PLS 100 POINTS TAKING TEST RN

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Answer:

I think he answer is C but I could be wrong

Explanation:

brainliest plz

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2 years ago

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According to the directions on a can of frozen orange juice concentrate, 1 can of concentrate is to be mixed with 3 cans of wate

WARRIOR [948]

Answer:

Explanation:

Given:

1 can of concentrate requires 3 cans of water

Now,

Total ounces in 200 6-ounce cans = 1200 ounces

also,

for 1 can of concentrate requires 3 cans of water

thus,

for 12 ounces can water can required = 3 × 12 ounces = 36 ounces of cans

Thus,

total ounce of juice per can = 12 + 36 = 48 ounces per can

therefore,

the number of 12-ounce cans required are = $\frac{\textup{Ttoal ounces}}{\textup{Ounces per can}}$

= $\frac{\textup{1200}}{\textup{48}}$

the number of 12-ounce cans required are = 25

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3 years ago

Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor

deff fn [24]

Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is, $A=8.82\ cm^2=0.000882\ m^2$

Angle between the uniform magnetic field and the horizontal, $\theta=31$

Magnetic flux, $\phi=4\times 10^{-4}\ Wb$

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

$\phi=BA\ cos\theta$

$\theta$ is the angle between magnetic field and the area

Here, $\theta=90-31=59^{\circ}$

$B=\dfrac{\phi}{A\ cos\theta}$

$B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}$

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

7 0

3 years ago

Ball A 1.55kg moving right at 8.76 m/s makes a head-on collision with ball B (0.752 kg) moving left at 11.4 m/s. After, ball B m

Illusion [34]

1.15 m/s to the left (3 sig. fig.).

<h3>Explanation</h3>

Momentum is conserved between the two balls if they are not in contact with any other object. In other words,

$p_{\text{A,initial}} + p_{\text{B,initial}}=p_{\text{A,final}} + p_{\text{B,final}}$

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$ , where

$m$ stands for mass and
$v$ stands for velocity, which can take negative values.

Let the velocity of objects moving to the right be positive.

$m_\text{A} = 1.55\;\text{kg}$ ,
$m_\text{B} = 0.752\;\text{kg}$ .

Before the two balls collide:

$v_\text{A} = +8.76\;\text{m}\cdot\text{s}^{-1}$ ,
$v_\text{B} = -11.4\;\text{m}\cdot\text{s}^{-1}$ .

After the two balls collide:

$v_\text{A}$ needs to be found,
$v_\text{B} = +9.03\;\text{m}\cdot\text{s}^{-1}$ .

Again,

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$ ,

$1.55 \times (+8.76) + 0.752 \times (-11.4) = 1.55\;{\bf v_{\textbf{A,final}}} + 0.752 \times (+9.03)$ .

$v_{\text{A,final}} = \dfrac{1.55 \times (+8.76) + 0.752 \times (-11.4)-0.752 \times (+9.03)}{1.55} = -1.15\;\text{m}\cdot\text{s}^{-1}$ .

$v_{\text{A,final}}$ is negative? Don't panic. Recall that velocities to the right is considered positive. Accordingly, negative velocities are directed to the left.