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antoniya [11.8K]

2 years ago

7

An automobile is traveling at a constant 15 m/s, then it undergoes acceleration from that moment forward. Which statement best d

escribes the automobile's new motion?

1 answer:

creativ13 [48]2 years ago

7 0

Answer:

d

Explanation:

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The city is thinking about starting a program to add a new chemical to the public pool. Before they do, they want to find out if

Kitty [74]

The best and most correct answer among the choices provided by the question is the fourth choice.

The best people for advising is <span>the government agency that regulates these types of chemicals.</span>
I hope my answer has come to your help. God bless and have a nice day ahead!

4 0

2 years ago

An archer tests various arrowheads by shooting arrows at a pumpkin that is suspended from a tree branch by a rope, as shown to t

erik [133]

Answer:

Bounce 1 , pass 3, emb2

Explanation:

(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle. So it is Bounce1, pass3, emb2.

6 0

2 years ago

The Americium nucleus, 241 95 Am, decays to a Neptunium nucleus, 237 93 Np, by emitting an alpha particle of mass 4.00260 u and

saveliy_v [14]

Answer:

The mass of Neptunium is 237.054 u.

Explanation:

Given that,

Mass of Americium = 241.05682 u

Mass of alpha particle = 4.00260 u

The equation is,

$_{95}^{241}Am\rightarrow _{93}^{237}NP+\alpha\ particle+5.5 MeV$

Let the mass of Neptunium is m.

Since the mass remain same.

We need to calculate the mass of Neptunium

Using formula of mass

Mass of Neptunium = Mass of Americium -Mass of alpha particle

Put the value into the formula

$m=241.05682 -4.00260$

$m=237.054\ u$

Hence, The mass of Neptunium is 237.054 u.

8 0

2 years ago

This sphere is now connected by a long, thin conducting wire to another sphere of radius R2 that is several meters from the firs

alekssr [168]

Here is the full question

A metal sphere with Radius R₁ has a charge Q₁. Take the electric potential to be zero at an infinite distance from the sphere

a) What are the electric field and electric potential at the surface of the sphere?

This sphere is now connected by a long, thin conducting wire to another sphere of radius R₂ that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached:

b) what is the total charge on each sphere?

Assume that the amount of charge on the wire is much less than the charge on each sphere.

Answer:

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b)

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

Explanation:

Given that;

the radius of the sphere = R

The radius of the first sphere = $R_1$

The radius of the second sphere = $R_2$

Charge on the first sphere = $Q_1$

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b) From the question. before the part b question; we learnt that the first sphere is now connected to another sphere;

Now that the two sphere are joined . Charges flows from one to another until their potentials are equal.

As Such; We use $q_1 \ and \ q_2$ to represent their charges respectively

The potential on the surface of the first sphere;

$V_1 = \frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}$

The potential on the surface of the second sphere;

$V_2 = \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

$V_1=V_2$

∴

$\frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}= \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

Thus, we can say :

$\frac{q_1}{q_2}= \frac{R_1}{R_2}$

and $Q_1 = q_1 + q_2$

As such ;

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

3 0

2 years ago

As planets move toward the sun, they tend to: A. move at constant speed B. speed up C. burn up D. slow down

dolphi86 [110]

Answer:

B. speed up

Explanation:

The acceleration of an object due to a body's gravity is:

g = GM / r²

where G is the universal constant of gravitation,

M is the mass of the body,

and r is the distance from the body.

As a planet approaches the sun, r decreases. As r decreases, g increases.

8 0

2 years ago