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3 years ago

9

A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began

to accelerate?

1 answer:

Montano1993 [528]3 years ago

8 0

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 = 36.95 m/s

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______ is/are usually the best conductor/s. a. Metals c. Air b. Water d. Minerals Please select the best answer from the choices

nataly862011 [7]

Metals are the best conductors.

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3 years ago

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A thin spherical shell with radius R1 = 4.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm . Both

kakasveta [241]

Answer:

The right response will be "450 volts".

Explanation:

The given values are:

R1 = 4.00 cm

R2 = 6.00 cm

q1 = +6.00 nC

q2 = −9.00 nC

As we know,

The potential difference between the two shell's difference will be:

⇒ $\Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]$

$=K[\frac{q1}{R2}-\frac{q1}{R1} ]$

On substituting the values, we get

$=(9\times 10^9)[\frac{6\times 10^{-9}}{0.04}-\frac{6\times 10^{-9}}{0.06}]$

Δ $=450 \ volts$

3 0

3 years ago

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

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3 years ago

How know that rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the arkose sandstone

svp [43]

The rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the Arkose sandstone because the Amadeus basin were made up of marine and non-marine sedimentary rocks which are softer compared to quarts which make up mostly the Arkose sandstone.

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3 years ago

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A has the magnitude 14.4 m and is angled 51.6° counterclockwise from the positive direction of the x axis of an xy coordinate sy

Ad libitum [116K]

Answer:

Ã in unit vector notation = 12.26485i + 7.54539j

B in unit vector notation = 16.3516i + 3.11529j

Explanation:

The detailed steps and calculation is shown in the attachment.

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3 years ago