Ask question

Ask question

All categories

3 years ago

5

Starting at 1.3 m/s, a runner accelerates at a constant 0.22 m/s2 for 6.0 s. What is the runner’s displacement during this time

interval?

1 answer:

Anettt [7]3 years ago

7 0

Answer:

answer is 11.76 meter

Explanation:

use 2nd equation of motion

S=ut+1/2at^2

You might be interested in

which statement is a Hypothesis? A: Most of the earthworms moved to the shaded area during experiment. B: If an earthworm is giv

True [87]

Answer:

B

Explanation:

It's saying what you think

3 0

3 years ago

What is the net force on a car moving in a straight line with a constant velocity

guapka [62]

Answer:

For example, when a car travels at a constant speed, the driving force from the engine is balanced by resistive forces such as air resistance and friction in the car's moving parts. The resultant force on the car is zero.

Explanation:

hope this helps

3 0

2 years ago

Read 2 more answers

A dockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal surface, requires a 73-N horiz

Nataliya [291]

1) Static friction coefficient: 0.355

The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.

The maximum static frictional force is given by

$F_s = \mu_s mg$

where

$\mu_s$ is the static coefficient of friction

m = 21 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have

$F_s=73 N$

Using this information into the previous equation, we can find the coefficient of static friction:

$\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355$

2) Kinetic friction coefficient: 0.267

Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is

$F_k = \mu_k mg$ (1)

where

$\mu_k$ is the coefficient of kinetic friction

There is a horizontal force of

F = 55 N

pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.

So we can write Newton's second law as

$F-F_k = ma = 0$

And by substituting (1), we can find the value of the coefficient of kinetic friction:

$F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267$

5 0

3 years ago

Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m2 to that for a double-pane window with

Rudik [331]

Answer:

2.80321285141

Explanation:

$L_g$ = Thickness of glass = 4.5 mm

$k_g$ = Thermal conductivity of glass = 0.8 W/mK

$R_0$ = Combined thermal resistance = $0.15\times m^2K/W$

$L_a$ = Thickness of air = 6.6 mm

$k_a$ = Thermal conductivity of air = 0.024 W/mK

The required ratio is the inverse of total thermal resistance

$\dfrac{2(L_g/k_g)+R_0+(L_a/k_a)}{(L_g/k_g)+R_0}\\ =\dfrac{2(4.5\times 10^{-3}/0.8)+0.15+(6.6\times 10^{-3}/0.024)}{(4.5\times 10^{-3}/0.8)+0.15}\\ =2.80321285141$

The ratio is 2.80321285141

4 0

3 years ago

Read 2 more answers

A wedge shaped air film is made between two sheets of glass using a spacer at one end of the glass sheets. If light of wavelengt

pychu [463]

Answer:

1178 nm

Explanation:

We are given that

Wavelength of light= $\lambda=589nm=589\times 10^{-9} m$

$1nm=10^{-9} m$

We have to find the thickness of spacer if five dark fringes are observed between the edges of the glass.

Suppose that first dark fringe and fifth dark fringe near spacer, then the path length of light is 4 times the wavelength of light.

The light passes through air film is two times then the change in air film thickness from one edge to other is two times the wavelength of light.

Change in air film thickness from one edge to other edge is same as the thickness of spacer.

Therefore, thickness of spacer= $2\lambda$

Thickness of spacer= $2\times 589\times 10^{-9}$ m

Thickness of spacer=1178 nm

Hence, the thickness of spacer=1178 nm

8 0

3 years ago