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2 years ago

If Juan upgrades from a sports car to a large truck, what will happen to the

Physics

1 answer:

Brrunno [24]2 years ago

5 0

Answer:

The force must increase

Explanation:

According to newton's second law "force is the product of mass and acceleration".

Force = mass x acceleration

Now, the mass of the sports car is lesser compared to that of the truck. Therefore, to take both automobiles to the same speed, enough force must be applied by the engine of the truck.

There must be an increase in the force in order to make both automobiles attain the same speed.

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The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st

Aleonysh [2.5K]

Answer:

$a = 5.83 \times 10^{-4} m/s^2$

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = $1.50 m/s^2$

now we know that

$F = ma$

$F = 70(1.50) = 105 N$

now for the space station will be same as above force

$F = ma$

$105 = 1.8 \times 10^5 (a)$

$a = \frac{105}{1.8 \times 10^5}$

$a = 5.83 \times 10^{-4} m/s^2$

3 0

3 years ago

two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist

Novosadov [1.4K]

Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

For series combination,

$R_{eq}=R_1+R_2$

For parallel combination,

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

When 6 ohm and 3 ohm are in series,

$R_s=6+3\\\\R_s=9\ \Omega$

When 6 ohm and 3 ohm are in paralle,

$\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega$

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.

6 0

2 years ago

If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

marshall27 [118]

Answer:

$P(3600)=593.247W$

Explanation:

First, let's find the voltage through the resistor using ohm's law:

$V=IR=20*8=160V$

AC power as function of time can be calculated as:

$P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)$ (1)

Where:

$\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency$

Because of the problem doesn't give us additional information, let's assume:

$\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi$

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

$P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W$

5 0

3 years ago

Calculate the critical angle for light going from Glycerine to air.

Georgia [21]

The refractive index for glycerine is $n_g=1.473$ , while for air it is $n_a = 1.00$ .

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
$\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}$

6 0

2 years ago

True or False<br> further the sun, the longer the shadow

xz_007 [3.2K]

Answer: Yes the further the sun is away the longer the shadow is. At noon,the shadow is the shortest because its straight up above you. If this helps pls mark brainliest!

6 0

2 years ago

Read 2 more answers