If you want to increase the acceleration of a baseball when you throw it, what must you increase to make this happen?

Two concrete spans of a 380 m long bridge are

Mazyrski [523]

Answer:

4.163 m

Explanation:

Since the length of the bridge is

L = 380 m

And the bridge consists of 2 spans, the initial length of each span is

$L_i = \frac{L}{2}=\frac{380}{2}=190 m$

Due to the increase in temperature, the length of each span increases according to:

$L_f = L_i(1+ \alpha \Delta T)$

where

$L_i = 190 m$ is the initial length of one span

$\alpha =1.2\cdot 10^{-5} ^{\circ}C^{-1}$ is the temperature coefficient of thermal expansion

$\Delta T=20^{\circ}C$ is the increase in temperature

Substituting,

$L_f=(190)(1+(1.2\cdot 10^{-5})(20))=190.0456 m$

By using Pythagorean's theorem, we can find by how much the height of each span rises due to this thermal expansion (in fact, the new length corresponds to the hypothenuse of a right triangle, in which the base is the original length of the spand, and the rise in heigth is the other side); so we find:

$h=\sqrt{L_f^2-L_i^2}=\sqrt{(190.0456)^2-(190)^2}=4.163 m$

4 0

4 years ago

A 10.0-kg mass is placed on a 25.0o incline and friction keeps it from sliding. The coefficient of static friction in this case

vovangra [49]

The frictional force while the mass is sliding will be 46.2 N.

<h3>What is friction force?</h3>

Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.

Given data:

m(mass)= 10.0-kg

Θ (Inclination angle)=25.0o

Coefficient of sliding friction, $\rm \mu_k$ =0.520

Coefficient of static friction, $\rm \mu_s=0.520$

The friction force, F=?

Resolve the force in the inclined plane;

$\rm F=\mu_s mg cos25^0 \\\\ F=0.520 \times 10 \times 9.81 \times cos 25 ^0 \\\\ F= 46.2 \ N$

Hence, the frictional force while the mass is sliding will be 46.2 N.

To know more about friction force refer to the link;

brainly.com/question/1714663

#SPJ1

5 0

2 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

4 years ago

The legal tradition that kept women from owning property and holding public office came to the United States from

scoundrel [369]

<span>The legal tradition that kept women from owning property and holding public office came to the United States from: C. Britain.</span>

6 0

4 years ago

Read 2 more answers

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$