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2 years ago

A car traveling at 100km/hr .how many hours will it take to cover a distance of 750 km

Physics

2 answers:

grin007 [14]2 years ago

4 0

Answer:

7.5 right?

Explanation: if im wrong shoot me

DedPeter [7]2 years ago

4 0

Answer:

7.5 hours( 7 hours 30 minutes)

Explanation:

Speed = distance/time

time = distance/speed

750 km divided by 100 km/hr = 7.5 hours

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One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago

A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/

Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly $y = 0.0204 \ m$

Explanation:

From the question we are told that

The diameter of the glass tube is $d = 1 \ mm = 0.001 \ m$

The density of glycerin is $\rho = 1260 \ kg /m^3$

The surface tension of the glycerin is $\sigma = 6.3 *10^{-2} \ N /m$

The capillary rise of the glycerin is mathematically represented as

$y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}$

substituting value

$y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}$

$y = 0.0204 \ m$

Therefore the height of the glass tube the glycerin was able to cover is

$y = 0.0204 \ m$

4 0

3 years ago

1.9 km converted to miles

Snowcat [4.5K]

When you convet km to miles this is what you get

1.18061

7 0

2 years ago

Bagaimana cara untuk melatih kemampuan melempar tangkap yang baik dalam bola basket

anyanavicka [17]

Answer:

pergi ke pertandingan sepak bola dan amati

7 0

2 years ago

You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the

lana [24]

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

$\Delta p=p_2-p_1$

The momentum is computed as

$p=mv$

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

$p_1=mv$

When it hits the wall, it sticks, thus its final speed is 0 and

$p_2=0$

The change of momentum is

$\Delta p=0-mv=-mv$

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

$p_2=-mv$

The change of momentum is

$\Delta p=-mv-mv=-2mv$

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0

3 years ago