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Nataly [62]
3 years ago
13

A car weighing 11.1 kN and traveling at 13.4 m/s without negative lift attempts to round an unbanked curve with a radius of 61.0

m. (a) What magnitude of the frictional force on the tires required to keep the car f static friction between the tire or not ("yes" or "no")? on its circular If the coefficient ro s 0.35, is the attempt at taking the curve successf (a) Number Units (b)
Physics
1 answer:
klasskru [66]3 years ago
8 0

Answer:

Well..

Explanation:

That's impossible. I know because I once weighed 11.1 kN, and I was temporarily immobile. It's probably the same for a car, and therefore it can not be "traveling" anywhere at all.. unless you put the car on an airplane or a boat or something.

You might be interested in
8. The legs of a young man are each 0.650 meters long. What is his maximum walking speed?
Norma-Jean [14]

Answer:

2.52 m/s

Explanation:

When the man takes a step, his foot is stationary while his body revolves around it.  At the point when his body is directly above his foot, there will be no normal force at his maximum speed.

Sum of the forces in the radial direction:

∑F = ma

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 0.650 m:

v = √(9.8 m/s² × 0.650 m)

v = 2.52 m/s

8 0
3 years ago
Letícia leaves the grocery store and walks 150.0 m to the parking lot. Then, she turns 90° to the right and walks an additional
Alexxx [7]

Answer:

165.529454

Explanation:

According to the Pythagorean Theorem for calculating the lengths of a right angle triangle's sides, a^2 + b+2 = c^2, where c is the longest side (and the side opposing the right angle). So in your case it would be 150*150 + 70*70 = 27400. And √ 27400 is your answer.

7 0
3 years ago
The "Biltmore Agreement" stipulated that a. remote broadcasts could not emanate from hotels. b. NBC could not give away tickets
Anna35 [415]

Answer:

The "Biltmore Agreement" stipulated that:

Radio stations agreed to broadcast no longer than five minutes of news, twice per day, while using information supplied by the newspapers.

e. radio stations could only air five-minutes newscasts a day.

Explanation:

The Biltmore Agreement tried to reconcile within the press war between newspapers and radio, as during its golden age the newspapers´ revenues decreased. Radio´s brand new technology was more attractive and creative for advertising and could report breaking news faster than the newspapers, which through the press associations including the Associated Press and the United Press, pressured to stop providing news to radio stations beginning a war in 1933, which partially ended with the Biltmore Agreement, which restricted the radio´s broadcasting of news if the newspapers continued publishing radio listings, radio stations were to broadcast no longer than five minutes of news, twice per day, if information supplied by the newspapers was used, no sponsors were allowed, and no more that 30 words in a single story were allowed either;  radio stations had to include: "See your daily newspaper for further details" in their announcements and, could only broadcast news after 9:30 AM for morning news, and after 9:00 PM for evening news, so people would have already received their newspapers.

7 0
3 years ago
Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and abov
makvit [3.9K]

Answer:

The mass of the solid is 16 units.

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

Work:

Density function: ρ(x, y, z) = 8

x-bounds: [0, 1], y-bounds: [0, x], z-bounds: [0, x+y+3]

The mass M of the solid is given by:

M = ∫∫∫ρ(dV) = ∫∫∫ρ(dx)(dy)(dz) = ∫∫∫8(dx)(dy)(dz)

First integrate with respect to z:

∫∫8z(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x+8y+24](dx)(dy)

Then integrate with respect to y:

∫[8xy+4y²+24y]dx, evaluate y from 0 to x

= ∫[8x²+4x²+24x]dx

Finally integrate with respect to x:

[8x³/3+4x³/3+12x²], evaluate x from 0 to 1

= 8/3+4/3+12

= 16

The mass of the solid is 16 units.

Now we have to find the center of mass of the solid which requires calculating the center of mass in the x, y, and z dimensions.

The z-coordinate of the center of mass Z is given by:

Z = (1/M)∫∫∫ρz(dV) = (1/16)∫∫∫8z(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫4z²(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[4(x+y+3)²](dx)(dy)

= ∫∫[4x²+24x+8xy+4y²+24y+36](dx)(dy)

Then integrate with respect to y:

∫[4x²y+24xy+4xy²+4y³/3+12y²+36y]dx, evaluate y from 0 to x

= ∫[28x³/3+36x²+36x]dx

Finally integrate with respect to x:

[7x⁴/3+12x³+18x²], evaluate x from 0 to 1

= 7/3+12+18

Z = (7/3+12+18)/16 = <u>2.021</u>

The y-coordinate of the center of mass Y is given by:

Y = (1/M)∫∫∫ρy(dV) = (1/16)∫∫∫8y(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8yz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8xy+8y²+24y](dx)(dy)

Then integrate with respect to y:

∫[4xy²+8y³/3+12y²]dx, evaluate y from 0 to x

= ∫[20x³/3+12x²]dx

Finally integrate with respect to x:

[5x⁴/3+4x³], evaluate x from 0 to 1

= 5/3+4

Y = (5/3+4)/16 = <u>0.3542</u>

<u />

The x-coordinate of the center of mass X is given by:

X = (1/M)∫∫∫ρx(dV) = (1/16)∫∫∫8x(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8xz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x²+8xy+24x](dx)(dy)

Then integrate with respect to y:

∫[8x²y+4xy²+24xy]dx, evaluate y from 0 to x

= ∫[12x³+24x²]dx

Finally integrate with respect to x:

[3x⁴+8x³], evaluate x from 0 to 1

= 3+8 = 11

X = 11/16 = <u>0.6875</u>

<u />

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

4 0
3 years ago
An object is thrown with velocity v from the edge of a cliff above level ground. Neglect air resistance. In order for the object
musickatia [10]

Answer:

(C) greater than zero but less than 45° above the horizontal

Explanation:

The range of a projectile is given by R = v²sin2θ/g.

For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°

2θ = 90°

θ = 90°/2 = 45°

So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.

4 0
3 years ago
Read 2 more answers
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