Ask question

Ask question

All categories

3 years ago

9

a car is moving with a velocity of 25m/s for 15s. calculate the displacement of the car. The acceleration of the car over the 15

s

1 answer:

SpyIntel [72]3 years ago

8 0

Explanation:

this might help you i think so

You might be interested in

Leto [7]

True
False
True
My answers

4 0

3 years ago

The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li

melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate $Q$ of a hypodermic needle:

$Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}$ (1)

Where the dimensions of each one is:

Volume flow rate $Q=\frac{L^{3}}{T}$

Radius of the needle $R=L$

Length of the needle $L=L$

Pressures at opposite ends of the needle $P_{2}$ and $P_{1}=\frac{M}{LT^{2}}$

Viscosity of the liquid $\eta=\frac{M}{LT}$

We need to find the value of $n$ whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

$\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}$ (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}$ (3)

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}$ (4)

As we can see $n$ must be 4 if we want the exponent to be 3:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}$ (5)

Finally:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}$ (6)

8 0

3 years ago

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

Anestetic [448]

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

5 0

3 years ago

a linear function has the same y-intercept as x + 4y equals 16 and it's graph contains the point (4,5). Find the slope of the li

navik [9.2K]

Answer: $\bold{\text{Slope (m)}=\dfrac{1}{4}}$

<u>Explanation:</u>

A linear equation is of the form: y = mx + b where

m is the slope
b is the y-intercept (where it crosses the y-axis)

x + 4y = 16

4y = -x + 16

$y = -\dfrac{1}{4}x+\dfrac{16}{4}$

$y=-\dfrac{1}{4}x+4$

The y-intercept (b) = 4

Next, find the slope given point (4, 5) and b = 4

$y=mx+b\\\\5=m(4)+4\\\\1=4m\\\\\dfrac{1}{4}=m\\\\\\\\\large\boxed{Slope (m)=\dfrac{1}{4}}$

6 0

3 years ago

Consider a lawnmower of mass m which can slide across a horizontal surface with a coefficient of friction μ. In this problem the

inna [77]

Answer:

Fh = u*m*g / (cos(θ) - u*sin(θ))

Explanation:

Given:

- The mass of lawnmower = m

- The angle the handle makes with the horizontal = θ

- The force applied along the handle = Fh

- The coefficient of friction of the lawnmower with ground = u

Find:

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Solution:

- Construct a Free Body Diagram (FBD) for the lawnmower.

- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:

F_net,x = m*a

- Since, the lawnmower is to move with constant speed then we have a = 0.

F_net,x = 0

- The forces as follows:

Fh*cos(θ) - Ff = 0

Where, Ff is the frictional force:

Fh = Ff /cos(θ)

Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:

- W - Fh*sin(θ) + Fn = 0

Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:

Fn = W + Fh*sin(θ)

Fn = m*g + Fh*sin(θ)

The Frictional force Ff is proportional to the contact force Fn by:

Ff = u*Fn

Ff = u*(m*g + Fh*sin(θ))

Substitute this expression in the form derived for Fh and Ff:

Fh*cos(θ) = u*(m*g + Fh*sin(θ))

Fh*(cos(θ) - u*sin(θ)) = u*m*g

Fh = u*m*g / (cos(θ) - u*sin(θ))

5 0

3 years ago