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3 years ago

ASAP answer all pls will mark brainiest

Physics

2 answers:

sammy [17]3 years ago

7 0

Answer:

1) B

2)D

Explanation:

I am unsure of #2

Brilliant_brown [7]3 years ago

5 0

1)C momentum is definitely a property just like inertia
2)D momentum = mass x energy = 9 x 3 = 27

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Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second

Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

5 0

3 years ago

A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m

Nina [5.8K]

Answer: $0.05\ mm$

Explanation:

Given

Cross-sectional area of wire $A_1=4\ mm^2$

Extension of wire $\delta l=0.1\ mm$

Extension in a wire is given by

$\Rightarrow \delta l=\dfrac{FL}{AE}$

where, $E=\text{Youngs modulus}$

$\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)$

for same force, length and material

$\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)$

Divide (i) and (ii)

$\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm$

5 0

3 years ago

3. Take sugar, oil, corn syrup, a glass and water. Pour the water in the glass and then add each of the above the substances one

hoa [83]

Here are the observations

Sugar:-

Sugar is soluble in water
so It will dissolve in water .

Corn syrup:-

Corn syrup is also basically a sugar.
It will dissolve in water too .
If we shake the mixture in glass then corn syrup will be dissolved.

Oil:-

Oil is not soluble in water
Hence it won't dissolve in water.
It will float over water and make two layers

7 0

3 years ago

Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

8 0

3 years ago

The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False

madreJ [45]

True

In fact, the weight of an object on the surface of the Earth is given by:
$F=mg$
where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
$F=mg=(3.0 kg)(9.81 m/s^2)=29 N$

8 0

3 years ago