Find the volume of an object with a mass of 63 grams and a density of 9 g/cm^3

Proposed Kinematic Exercise I

valentinak56 [21]

Answer:

(a) 20 m

(b) 6 m/s²

(c) Between t=0 and t=2, the body moves to the left.

Between t=2 and t=4, the body moves to the right.

Explanation:

v = 3t² − 6t

x(0) = 4

(a) Position is the integral of velocity.

x = ∫ v dt

x = ∫ (3t² − 6t) dt

x = t³ − 3t² + C

Use initial condition to find value of C.

4 = 0³ − 3(0)² + C

4 = C

x = t³ − 3t² + 4

Find position at t = 4.

x = 4³ − 3(4)² + 4

x = 20

(b) Acceleration is the derivative of velocity.

a = dv/dt

a = 6t − 6

Find acceleration at t = 2.

a = 6(2) − 6

a = 6

(c) v = 3t² − 6t

v = 3t (t − 2)

The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.

When 0 < t < 2, v < 0.

When t > 2, v > 0.

3 0

4 years ago

Describe what a convection current is in how it was material than the earth

Leona [35]

The movement of fluid as a result of differential heating or convection. Earth convection currents refer to the motion of molten rock in the mantle as radioactive decay heats up magma, causing it to rise and driving the global scale flow of magma.

3 0

3 years ago

Why must chemical rxn be balanced

Svetradugi [14.3K]

The chemical equation needs to be balanced so that it follows the law of conservation of mass. A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side. Balancing chemical equations is a process of trial and error.

5 0

3 years ago

The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i

attashe74 [19]

Let the vector position of the object in the (x-y) plane be
$\vec{r} = x \hat{i} + y \hat{j}$

The applied force is
$\vec{f} = -12.5 \hat{i}
$

By definition, the applied torque is
$\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}$

Answer: $12.5y \, \hat{k}$

7 0

4 years ago

Suppose a convex mirror has a focal length of 120 cm. A candle sits directly in front of the mirror. If the image of that candle

Andru [333]

We can solve the problem by using the mirror equation:
$\frac{1}{f} = \frac{1}{d_o}+ \frac{1}{d_i}$
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

For the sign convention, the focal length is taken as negative for a convex mirror:
$f=-120 cm$
and the image is behind the mirror, so virtual, therefore its sign is negative as well:
$d_i=-24 cm$
putting the numbers in the mirror equation, we find the distance of the object from the mirror surface:
$\frac{1}{d_o} = \frac{1}{f}- \frac{1}{d_i}= \frac{1}{-120 cm} - \frac{1}{-24 cm}= \frac{1}{30 cm}$
So, the distance of the object from the mirror is $d_o = 30 cm$

8 0

3 years ago