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3 years ago

A box is being pushed by two stellar science students, one on each side of the box. Delfino is pushing

Physics

1 answer:

Anettt [7]3 years ago

6 0

Explanation:

It is given that,

Delfino is pushing the box with a force of 10 N to the left. Aleida is pushing the box with a force of 15 N to the right.

Aleida is exerting more force. He is the more stronger individual.

Net force is given by :

F = -10+15

F = 5 N

So, the net force is 5 N and it is in right direction.

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A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

6 0

4 years ago

Write a rule for the sequence. 3, -3, -9, -15. A. Start with 3 and add -6 repeatedly B. Start with -6 and add 3 repeatedly C. St

faust18 [17]

Substract two consecutive terms of the sequence to see if there is a common difference:

$\begin{gathered} (-3)-(3)=-3-3=-6 \\ (-9)-(-3)=-9+3=-6 \\ (-15)-(-9)=-15+9=-6 \end{gathered}$

As we can see, there is a common difference of -6.

Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).

Notice that the first term of the sequence is 3.

Then, the rule for the sequence is to start with 3 and add -6 repeatedly.

Therefore, the correct choice is option A) Start with 3 and add -6 repeatedly.

7 0

2 years ago

A major difference between radio waves, visible light, and gamma rays is the ____ of the photons, which results in the different

olga2289 [7]

The sentence can be completed as follows:
"A major difference between radio waves, visible light, and gamma rays is the energy of the photons, which results in the different photon frequencies and wavelengths."

In fact, gamma rays have greater energy than visible light and visible light has greater energy than radio waves. The energy E of a photon is related to its frequency, f, by
 $E=hf$
where h is the Planck constant. We see from this formula that the higher the frequency, the greater the energy. Instead, the wavelength is inversely proportional to the frequency:
 $\lambda= \frac{c}{f}$
where c is the speed of light. Since the frequency is directly proportional to the energy, this means that the wavelength is inversely proportional to the energy.

8 0

3 years ago

Read 2 more answers

Energy needs, in total kilocalories per day, are greater during _____________ than any other time of life.

malfutka [58]

Energy needs, in total kilocalories per day, are greater during adolescence than any other time of life. The correct option among all the options that are given in the question is the second option or option "b". Only during the time of pregnancy will a girl require more kilocalories per day than during the time of adolescence.

5 0

3 years ago

Read 2 more answers

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

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

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago