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dlinn [17]
3 years ago
12

The wind pushes a paper cub along the sand at a beach. The cup has a mass of 25 grams (= ? Kg's) and accelerates at a rate of 5

m/s/s. How much force (in Newtons) is the wind exerting on the cup
Physics
1 answer:
bixtya [17]3 years ago
7 0

Force = (mass) x (acceleration)

= (0.025 kg) x (5 m/s²)

=   0.125 Newton
You might be interested in
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 n
hichkok12 [17]

Answer:

891 excess electrons must be present on each sphere

Explanation:

One Charge = q1 = q

Force = F = 4.57*10^-21 N  

Other charge = q2 =q

Distance = r = 20 cm = 0.2 m  

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)  

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q =  r*sq rt F[4pieo]

 q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]

q = 1.42614*10^ -16 C

number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19

n =891

 891 excess electrons must be present on each sphere  

5 0
3 years ago
Asteroids, meteoroids, and comets are remnants of the early solar system. (T/F)
4vir4ik [10]

Answer: Asteroids, meteoroids, and comets are remnants of the early solar system. This Statement is TRUE.

Explanation:

METEOROID: these are small rocky or metallic objects found in outer space.

ASTEROIDS: these are also known as minor planets of the inner solar system. They are irregularly shaped object in space that orbits the Sun.

COMETS: these are dusty chunk of ice, that moves in a highly elliptical orbit about the sun.

Asteroids, meteoroids, and comets as remnants of the early solar system was further proved in nebular hypothesis

initially proposed in the eighteenth century by German philosopher Immanuel Kant and French mathematician Pierre-Simon Laplace. (The word nebula means a gaseous cloud.) According to the modern version of the theory, about 4.5 to 5 billion years ago the solar system developed out of a huge cloud of gases and dust floating through space. These materials were at first very thin and highly dispersed.

8 0
3 years ago
The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0
3 years ago
he capacitor can withstand a peak voltage of 550 VV . If the voltage source operates at the resonance frequency, what maximum vo
anygoal [31]

Answer:

The maximum voltage is 39.08 V.

Explanation:

Given that,

Voltage = 550 V

Suppose, In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is 1.20\times10^{-2}\mu F

We need to calculate the resonant frequency

Using formula of resonant frequency

f=\dfrac{1}{2\pi\sqrt{LC}}

Put the value into the formula

f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}

f=2356.8\ Hz

We need to calculate the maximum current

Using formula of current

I=\dfrac{V_{c}}{X_{c}}

I=2\pi f C\times V_{c}

Put the value into the formula

I=2\pi\times2356.8\times1.20\times10^{-8}\times550

I=0.0977\ A

We need to calculate the impedance of the circuit

Using formula of impedance

Z=\sqrt{R^2+(X_{L}-X_{C})^2}

At resonant frequency , X_{L}=X_{C}

So, Z = R

We need to calculate the maximum voltage

Using formula of voltage

V=IR

Put the value into the formula

V=0.0977\times400

V=39.08\ V

Hence, The maximum voltage is 39.08 V.

4 0
3 years ago
A man walks 30m to the west, then 5m to the east in 45 seconds. What is the average velocity?
solong [7]
1.28m/s the velocity is found by distance/time.
8 0
3 years ago
Read 2 more answers
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