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Sever21 [200]
2 years ago
9

A car increases its speed from 4.20 m/s to 8.60 m/s over 3.20 seconds. What is the car’s acceleration?

Physics
1 answer:
katrin2010 [14]2 years ago
5 0

Answer:1.375metre per second square

Explanation: acceleration=(final velocity-initial velocity)÷time

acceleration=(8.6-4.2)÷3.2

Acceleration=4.4÷3.2

Acceleration=1.375 metre per second square

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Help please! This question is driving me crazy
Varvara68 [4.7K]

Answer:

-10.8°, or 10.8° below the +x axis

Explanation:

The x component of the resultant vector is:

x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)

x = 4.07

The y component of the resultant vector is:

y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)

y = -0.777

Therefore, the angle between the resultant vector and the +x axis is:

θ = atan(y / x)

θ = atan(-0.777 / 4.07)

θ = -10.8°

The angle is -10.8°, or 10.8° below the +x axis.

3 0
3 years ago
How does the gravity which pulls the moon and earth toward each other also affect the ocean?
slamgirl [31]
Gravity is what holds the planets in orbit around the sun and what keeps the moon in orbit around Earth. The gravitational pull of the moon pulls the seas towards it, causing the ocean tides. Gravity creates stars and planets by pulling together the material from which they are made.
5 0
3 years ago
Read 2 more answers
If the pressure exerted by the liquid of density 500kg/m3 is 20,000Pa. What is height of liquid column.
gregori [183]

The height of the liquid column is 4.08 metres.

4 0
2 years ago
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
2 years ago
Can you help me please the question says
bearhunter [10]

Answer:

p= 4 m/v

Explanation:

v=l*w*h

v=(25)(2)(3)

v=150

p=600/150

p=4

7 0
3 years ago
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