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3 years ago

A body travels 30m in 5s, 45m in 7s and then 65m in the last 5s. Find the average speed of the body

1 answer:

8 0

We can find the average speed of the body by finding the total distance covered, and then dividing it by the total time of the motion.

The total distance covered is:
$S=30 m + 45m+65m=140 m$

while the total time of the motion is
$t=5 s+7s+5 s=17 s$

So, the average speed of the body is:
$v= \frac{S}{t}= \frac{140 m}{17 s}=8.24 m/s$

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A proton moves in the negative x-direction through a uniform magnetic field in the negative y-direction what is the direction of

ANEK [815]

A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.

<h3>What is the right-hand thumb rule?</h3>

Hold a current-carrying conductor in your right hand with your thumb pointing in the direction of the current then wrap your fingers around the conductor and orient them in the direction of the magnetic field lines.

A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction.

The proton will be subject to a magnetic pull that is directed into the page.

Hence, option B is correct.

To learn more about the right-hand thumb rule refer to the link;

brainly.com/question/11521829

#SPJ1

4 0

2 years ago

alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;

zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0

3 years ago

Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a

Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

Force applied on camel feet = 4000 N
Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

${\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}$

Substituting all the given values in the formula to find the pressure exerted by camel feet.

$\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\ \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}} \\ \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\ \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\ \\\star \: \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}$

Hence, the pressure exerted by camel feet is 2000 N/m².

$\rule{300}{2.5}$

3 0

2 years ago

Which of the following is a balanced chemical equation?

Natalija [7]

Explanation:e=mwS

<em>and also yes it is i think </em>

5 0

3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work

aksik [14]

It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
$F=k_e \frac{q_e q_e}{d^2}$
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.

5 0

3 years ago