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3 years ago

(1) Define uniform acceleration

Physics

2 answers:

marissa [1.9K]3 years ago

8 0

Explanation:

If an object's speed (velocity) is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration.

A body has a uniform acceleration if it travels in a straight line and its velocity increases by equal amount in equal intervals of time. For Ex:Motion of freely falling body, bicycle going down the slope, ball rolling down in inclined plane.

Anettt [7]3 years ago

6 0

Answer:

Explanation:

When an object's speed increases at a constant rate, we say it has constant/uniform acceleration.

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A river has a steady speed of 0.550 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point. If

nlexa [21]

Answer:

$t=564.83seconds=9.414minutes=0.1569hours$

Explanation:

Here the steady speed of the river water relative to ground is

$V_{wg}=0.55m/s$

Speed of the boy relative to still water is

$V_{BW}=1.50m/s$

Here we are considering +ve speed along along +ve x-axis and -ve speed along -ve x-axis

Therefore the speed of boy relative to the ground upstream is:

$V_{BGup}=V_{BW}-V_{WG}\\V_{BGup}=1.50m/s-0.550m/s\\V_{BGup}=0.95m/s$

And the speed of boy relative to the ground downstream is:

$V_{BGdown}=V_{BW}+V_{WG}\\V_{BGdown}=1.50m/s+0.550m/s\\V_{BGdown}=2.05m/s$

The distance covered in upstream trip d₁=1.0km=1000m and the distance covered in downstream is d₂=1.0km=1000m

Since time taken by a person is to cover distance d with speed v given by:

$t=d/v$

The total time taken for one trip is:

$t=\frac{d_{1} }{V_{BGup} } -\frac{d_{2} }{V_{BGdown} }\\t=\frac{1000m}{0.95m/s}-\frac{1000m}{2.05m/s}\\ t=564.83seconds=9.414minutes=0.1569hours\\$

3 0

3 years ago

What kind of waves can travel through empty space or a vacuum?

satela [25.4K]

Answer:

the answer is B.

4 0

4 years ago

What is the value of a conversion factor ratio? A.1 B.3 C.10 D.12

Alecsey [184]

A is the correct answer :D

4 0

4 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

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4 years ago

Use the SI prefixes in Table 3 of this chapter to convert these hypothetical units of measure into appropriate quantities: a. 10

Amanda [17]

The first response would be "deca," since this is a multiple of 10^1. B, since it's working off the thousands prefix (10^3), would be "kilo." The third, at 10^-6, would be "micro." Next, at 10^-9, would be "nano," and the final, 10^18, would have a prefix of "exa."

6 0

4 years ago