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3 years ago

A car goes from rest to a speed of 90 km/h in 10 seconds . What is the car's acceleration in m/s²

1 answer:

7 0

First you must convert Km/hr to m/s. 90 km/hr equals 25m/s (this can be done through a conversion table by plugging in the conversion values). Then you need to see what was given:
vi (initial velocity)= 0m/s
vf (final velocity= 25m/s (90km/hr)
t (time)= 10s

Next you should find an equation that requires only the values you know and gives you the value you're looking for. Sometimes that requires two equations to be used, but in this case you only need one. The best equation for this would be a=(vf-vi)/t. Finally, plug in your values (a=(25-0)/10) to get your answer which would be 2.5m/s^2. Hope this helped!

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"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after

wel

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as

K =-V dP/dV

Where K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x 4 × 106N/m2 / dV

dV = - 4,300 x 4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

3 0

3 years ago

During which phase of the moon do neap tides occur?

Fynjy0 [20]

Answer:

First Quarter and Third Quarter.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.

Since gravity variates with the distance:

$F = G\frac{m1\cdot m2}{r^{2}}$ (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).

However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.

Therefore, that happens when the Moon is at First Quarter and Third Quarter.

4 0

3 years ago

Read 2 more answers

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

7 0

3 years ago

Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

$E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s$

The velocity of the flea when leaving the ground is 1.327363 m/s

$W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m$

The flea will travel 0.00090243 m upward

8 0

3 years ago

A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.

katrin2010 [14]

Answer:

The correct answer is a

Explanation:

At projectile launch speeds are

X axis vₓ = v₀ = cte

Y axis $v_{y}$ = v_{oy} –gt

The moment is defined as

p = mv

For the x axis

pₓ = mvₓ = m v₀ₓ

As the speed is constant the moment is constant

For the y axis

p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)

Speed changes over time, so the moment also changes over time

Let's examine the answer

i True

ii False. The moment changes with time

The correct answer is a

7 0

3 years ago