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Genrish500 [490]

2 years ago

10

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

po alcanzará el coche a la furgoneta? ¿A qué distancia se producirá el encuentro?

1 answer:

statuscvo [17]2 years ago

6 0

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)

xo = 10 km

v1 = 55km/h

car 2:

$x'=v_2t$ (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

$x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}$

$t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s$

The position in which both cars coincides is:

$x=(55km/h)(0.33h)=18.33km$

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Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. The electric field at a point P, which

krok68 [10]

Answer:

$E_{max}=41666.66\ N/C$

Explanation:

Given that,

The radius of sphere, r = 0.3 m

Distance from the center of the sphere to the point P, x = 0.5 m

Electric field at point P, $E_P=15000\ N/C$ (radially outward)

The maximum electric field is at the surface of the sphere. We know that the electric field is inversely proportional to the distance. So,

$\dfrac{E_{max}}{E_p}=\dfrac{0.5^2}{0.3^2}$

$\dfrac{E_{max}}{15000}=\dfrac{0.5^2}{0.3^2}$

${E_{max}}=\dfrac{0.5^2}{0.3^2}\times 15000$

$E_{max}=41666.66\ N/C$

So, the magnitude of the electric field due to this sphere is 41666.66 N/C. Hence, this is the required solution.

6 0

2 years ago

Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of

zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, $m=40\ kg$

Radius of the barrel is, $R=2.90\ m$

Number of revolutions are, $n =10$

Time taken for 10 revolutions is, $t=29.3\ s$

Therefore, the time period of the child is given as:

$T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s$

Now, angular velocity is related to time period as:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s$

Now, centripetal force acting on the child is given as:

$F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N$

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0

2 years ago

The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

SashulF [63]

Answer:

7.9 $\frac{m}{s^{2} }$

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

$\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\$

Using algebra, we can rearrange our equation to find the final acceleration, $a_{2}$ :

$a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\$

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 $\frac{m }{s^{2} }$ = 8.33 $\frac{m }{s^{2} }$

Plug everything in:

7.9 $\frac{m }{s^{2} }$ = $\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}$

(1590kg the initial weight plus the weight of the added passenger)

8 0

2 years ago

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T. If the magnetic

larisa86 [58]

Answer:

The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

Explanation:

Given that,

Charge $q =8.4\times10^{-4}\ C$

Angle = 35°

Magnetic field strength $B=6.7\times10^{-3}\ T$

Magnetic force $F=3.5\times10^{-2}\ N$

We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

$F = qvB\sin\theta$

$v = \dfrac{F}{qB\sin\theta}$

Where,

F = Magnetic force

q = charge

B = Magnetic field strength

v = velocity

Put the value into the formula

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times\sin35^{\circ}}$

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times0.57}$

$v = 10910.36\ m/s$

$v = 1.1\times10^{4}\ m/s$

Hence, The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

4 0

2 years ago

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas i

asambeis [7]

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

5 0

2 years ago