Need help! 10 points

It took a student 30 minutes to drive from his home to campus on

Gennadij [26K]

Answer:

48 i believe

Explanation:

3 0

3 years ago

A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured

Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}$

Put the value into the formula

$0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}$

$v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}$

$v_{3}=4.98\ m/s$

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

6 0

2 years ago

Dry air will break down if the electric field exceeds about 3.0×106v/m. part a what amount of charge can be placed on a capacito

Genrish500 [490]

The solution for this problem is:The charge would be now equal to:(electric constant) multiplied by the (field strength) multiplied by the (area) so plugging in our values, will give us:8.85 * 10^-12 As / (V * m) * 3 * 10^6 V/m * 0.055 m^2 = 1.46 e-6 amperes would be the answer

7 0

3 years ago

Although the class has no water, the students want to know if the cube will float or sink in water. Explain, in detail, the step

Nutka1998 [239]

Okay, so the density of water is 1g/cm3. In order for the cube to float, it has to be less than 1, and it will sink if it is more than 1 g/cm3. Use a triple beam balance to weigh the cube, looking at the metric ruler on the balance. Then, if the cube's density is more than 1, then you know it will float. If the density is less than 1, you know it will sink.
hope this helps, and I didn't know how to use the word "metric ruler"

3 0

3 years ago

Imagine another solar system, with a star of the same mass as the Sun. Suppose a planet with a mass twice that of Earth (2 Earth

sashaice [31]

Answer:

Time period, T = 403.78 years

Explanation:

It is given that,

Orbital distance, $a=1\ AU=1.496\times 10^{11}\ m$

Mass of the Earth, $m_e=5.972\times 10^{24}\ kg$

Mass of the planet, $m_p=2m_e=11.944\times 10^{24}\ kg$

Let T is the orbital period of this planet. The Kepler's third law of motion gives the relation between the orbital period and the orbital distance.

$T^2=\dfrac{4\pi^2}{Gm_p}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 11.944\times 10^{24}}\times (1.496\times 10^{11})^3$

$T=1.28\times 10^{10}\ s$

or

T = 403.78 years

So, the orbital period of this planet is 404 years. Hence, this is the required solution.

4 0

2 years ago