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3 years ago

10). Three - fourth of the volume of a block which is floating in water extends above the surface

Physics

1 answer:

ikadub [295]3 years ago

5 0

Answer:

Explanation:

Fill the graduated cylinder partially with water to a level where you can submerge the object and drop the sinker weight into the water. If you don't have a graduated cylinder to fit your object, put a cylinder in a basin, fill it to the top with water and measure the overflow into the basin. Your answer will be less accurate because of the number of times the water has been moved. Note the amount of displacement in milliliters (ml) caused by the sinker and string.

Measure the mass of your object (say a cork) on a balance scale in grams (g). Be sure the object is dry when it is measured. Record its weight. Attach the sinker with the string to the object. If you use a staple or pin, be sure to include that when you measure the displacement of the sinker in step one.

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A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc

Hoochie [10]

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v= 0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

$a=- \dfrac{20^2}{2\times 30} \ m/s^2$

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be - 6.667 m/s² .

6 0

2 years ago

The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give

Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0

2 years ago

In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo

irinina [24]

Answer:

The torque about the origin is $2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}$

Explanation:

Torque $\overrightarrow{\tau}$ is the cross product between force $\overrightarrow{F}$ and vector position $\overrightarrow{r}$ respect a fixed point (in our case the origin):

$\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}$

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

$\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]$