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saul85 [17]
3 years ago
6

A box is being pulled to the right. What is the direction of the gravitational force?

Physics
2 answers:
lapo4ka [179]3 years ago
7 0

The correct answer to the question is vertically downward i.e towards the centre of earth.

EXPLANATION:

As per the question, the box is pulled to the right.

Hence, the direction of the applied force is towards right.

We are asked to determine the direction of the gravitational force that acts on the body.

Before answering this question, first we gave to understand the gravitational force of earth.

Any body present on the surface of earth is attracted with the force of gravity of earth ( gravitational force ) towards its centre.  It is equivalent to the weight of the body.  

The force of gravity is always directed towards the centre of earth irrespective of the nature of applied force.

Hence, the direction of the gravitational force which acts on the box is vertically downward.


dedylja [7]3 years ago
7 0
The direction of the gravitational force is always toward the center
of the Earth.  Wherever you are, it's the direction you call "down".
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Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is
Llana [10]

Answer:

K_e_q=22.75878093\frac{N}{m}

f=1.363684118Hz

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}

Replacing the data provided:

\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}

K_e_q=22.75878093\frac{N}{m}

Finally, to calculate the frequency of oscillation we use this:

f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }

Replacing m and k:

f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz

4 0
3 years ago
A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,
tamaranim1 [39]

Answer: a)  The rate constant, k, for this reaction is 0.00516s^{-1}

b) No t_{\frac{1}{2}} does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A\rightarrow products

Given: Order with respect to A = 1

Thus rate law is:

a) Rate=k[A]^1

k= rate constant

0.00250=k[0.484]^1

k=0.00516s^{-1}

The rate constant, k, for this reaction is 0.00516s^{-1}

b) Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}

t_{\frac{1}{2}}=\frac{0.69}{k}

Thus t_{\frac{1}{2}} does not depend on concentration.

8 0
3 years ago
A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms va
natima [27]

Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor C= 2.00\mu F

Current = 0.680 A

We need to calculate the inductance of the inductor

Using formula of current

I = \dfrac{V}{Z}

Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}

Put the value of Z into the formula

I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}

Put the value into the formula

0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}

L=35.8\ mH

Hence, The inductance of the inductor is 35.8 mH

4 0
3 years ago
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As temperature decreases, the steel will contract (volume decreases) thereby, an increase in Density.
3 0
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Tech A says that a vacuum modulator converts manifold vacuum into an engine load signal. Tech B says that the manual valve is mo
alexandr1967 [171]

Answer:

The correct option for the answer is A.) Tech A

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Tech A is correct

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