An endothermic reaction is defined as a reaction that ____

A 1400 kg car traveling at 17.0 m/s to the south collides with a 4700 kg truck that is at rest. The car and truck stick together

STatiana [176]

Answer:

Final velocity = 7.677 m/s

KE before crash = 202300 J

KE after crash = 182,702.62 J

Explanation:

We are given;

m1 = 1400 kg

m2 = 4700 kg

u1 = 17 m/s

u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J

8 0

3 years ago

Which transition represents a time when water molecules are moving closer together?

slava [35]

A. Is very attractive. If it's sublimation directly from water vapor in the air to ice on the glass, then yes. But from liquid water mist to water ice, no. Ice is less dense than water. That's why cubes float in your soda. Better leave 'A' alone. . . . D. Ice pellets turn to liquid. That one's good.

8 0

3 years ago

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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

3 0

3 years ago

Which statement BEST describes Miguel Hidalgo?

tatiyna

Miguel Hidalgo was a Priest so I think it would be 1.
Hope I Helped ( :

6 0

3 years ago

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$