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2 years ago

8

a car has a velocity of 10ms-1. it accelerates at 0.2 ms-2 for half minute find the total distance travelled and final velocity

of car.

1 answer:

Readme [11.4K]2 years ago

5 0

Answer:

Final velocity = 16 m/s

Total distane travelled = 390 m

Explanation:

We can use equation of motion to solve this:

$v = u + at \\ v = 10 + 0.2(30) \\ v = 16 {ms}^{ - 1}$

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 10(30) + \frac{1}{2} (0.2) {(30)}^{2} \\ s = 390m$

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Explain the benefits of understanding sound waves properties in real life.

Tems11 [23]

Answer:

The vibrations from sound waves cause our ears to send signals to our brains to create sound. The speed of sound waves will determine the sound's pitch, or how high or low something sounds. Sound waves are important because they allow us to hear important messages and emergency signals to protect ourselves.

Explanation:

I hope this helps :)

4 0

2 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago

A potential energy function is given by U(x)=(3.00J)x+(1.00J/m2)x3. What is the force function F(x) (in newtons) that is associa

Anit [1.1K]

Answer:

$F(x)=-3 N - (3N)x^2$

Explanation:

The force is defined as the negative of the derivative of the potential energy:

$F=-\frac{dU}{dx}$

If we use the potential energy function given in this problem:

$U(x)=3.00 x + 1.00 x^3$

and we calculate the force, we get:

$F(x)=-\frac{d}{dx}(3x+x^3)=-3-3x^2$

So, the force is

$F(x)=-3 N - (3N)x^2$

5 0

2 years ago

Read 2 more answers

A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th

ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a) Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0

2 years ago

A 1.30-kg object is held 1.10 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is droppe

pshichka [43]

Answer:

0.345m

Explanation:

Let x (m) be the length that the spring is compress. If we take the point where the spring is compressed as a reference point, then the distance from that point to point where the ball is held is x + 1.1 m.

And so the potential energy of the object at the held point is:

$E_p = mgh$

where m = 1.3 kg is the object mass, g = 10m/s2 is the gravitational acceleration and h = x + 1.1 m is the height of the object with respect to the reference point

$E_p = 1.3 * 10 * (x + 1.1) = 13(x + 1.1) = 13x + 14.3 J$

According to the conservation law of energy, this potential energy is converted to spring elastic energy once it's compressed

$E_p = E_k = kx^2/2 = 13x + 14.3$

where k = 315 is the spring constant and x is the compressed length

$315x^2 = 26x + 28.6$

$315x^2 - 26x - 28.6 = 0$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{26 \pm \sqrt{26^2 - 4*(-28.6)*315}}{2*315}$

$x = \frac{26 \pm 191.6}{630}$

x = 0.345 m or x = -0.263 m

Since x can only be positive we will pick the 0.345m

6 0

3 years ago