Ask question

Ask question

All categories

3 years ago

9

a typical top fuel dragstar has a mass of 900 kg what is the force applied to a dragstar by the static friction of the strip the

pushes it down road? f = ma

1 answer:

AnnZ [28]3 years ago

6 0

Wonderful.
You've given us ' m '.
Now, if we only had ' a ', we could come back at you with ' f '.

You might be interested in

Given the indices of refraction n_1 and n_2 of material 1 and material 2, respectively, rank these scenarios on the basis of the

aniked [119]

Answer:

a) order of refraction is a, e, de, c , c) a f g

Explanation:

a) when lightning is refracted it must comply with the law of refraction

.n₁ sin θ₁ = n₂ sinθ₂

sin θ / sin δδa) when lightning is refracted it must comply with the law of refraction

.n1 sin θ₁ = n2 sin θ ₂

Sint θ1 / sin θ2 = n2 / n1

1 we see that the ray deviation is promotional to the ratio of the refractive indices

The order of refraction is a, e, de, c

.b) When a ray passes from a medium with less indicated refraction to a higher index the part of the reflected ray has a phase change of 180º

a) no phase change

b) reflected ray has a phase change of 180º

c) no phase change

d) no phase change

e) there is a phase change

f) these phase changeo

1 we see that the ray deviation is promotional to the ratio of the refractive indices

The order of refraction is a, e, de, c

.b) When a ray passes from a medium with less indicated refraction to a higher index the part of the reflected ray has a phase change of 180º

a) no phase change

b) reflected ray has a phase change of 180º

c) no phase change

d) no phase change

e) there is a phase change

f) ay phase vabio

7 0

3 years ago

A refrigerator is 1.8m tall, lm wide,and 0.8m deep.The center of mass is lm from the bottom, 0.5m from the side, and 0.6m from t

VikaD [51]

Answer:

F = 520 N

Explanation:

For this exercise the rotational equilibrium equation should be used

Σ τ = 0

Let's set a reference system with the origin at the back of the refrigerator and the counterclockwise rotation as positive. On the x-axis it is horizontal directed outward, eg the horizontal y-axis directed to the side and the z-axis vertical

Torque is

τ = F x r

the bold indicate vectors, we analyze each force

the applied force is horizontal along the -x axis, the arm (perpendicular distance) is directed in the z axis,

The weight of the body is the vertical direction of the z-axis, so the arm is on the x-axis

-F z + W x = 0

F z = W x

F = $\frac{x}{z}$ W

The exercise indicates the point of application of the force z = 1.5 m and the weight is placed in the center of mass of the body x = 0.6 m, we are assuming that the force is applied in the wide center of the refrigerator

let's calculate

F = 1300 0.6 / 1.5

F = 520 N

5 0

3 years ago

Disadvantage of entrepreneurship

pochemuha

Answer:

limited liability, limitation in expansion, risk bearing, problem of continuity,

7 0

3 years ago

Describe the main distinguishing features of spiral, elliptical, and irregular galaxies.

hram777 [196]

Answer:

Spiral galaxies consist of a flat, rotating disk of stars, gas and dust, and a central concentration of stars known as the bulge. These are surrounded by a much fainter halo of stars, many of which reside in globular clusters.

Elliptical galaxies have smooth, featureless light-profiles and range in shape from nearly spherical to highly flattened, and in size from hundreds of millions to over one trillion stars. In the outer regions, many stars are grouped into globular clusters. Most elliptical galaxies are composed of older, low-mass stars, with a sparse interstellar medium and minimal star formation activity They are often chaotic in appearance, with neither a nuclear bulge nor any trace of spiral arm structure. Collectively they are thought to make up about a quarter of all galaxies.

irregular galaxies were once spiral or elliptical galaxies but were deformed by gravitational action. they are shapeless.

5 0

3 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

soldi70 [24.7K]

$Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}$

$GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} } \approx 1.42\cdot 10^{12}m$

3 0

3 years ago