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3 years ago

Tyler and his family drove from Atlanta to Baltimore. They traveled a total of 678 miles over 2 days.

Physics

2 answers:

sweet-ann [11.9K]3 years ago

7 0

Answer:

Atlanta

Explanation:

Keith_Richards [23]3 years ago

6 0

Answer:

atlanta

Explanation:

just took the test

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Calculate the speed of a bike rider who accelerates (from rest) for 5 seconds down a hill at an acceleration of 8

olya-2409 [2.1K]

Answer:

speed=abs(v)=40ms^-1

Explanation:

acceleration, a = (v-u)/t

since initial velocity u=0 (at rest) and a=8ms^-2,

8=v/5

hence after 5 seconds, v=40ms^-1

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3 years ago

The velocity - time graph of a ball of mass 25g moving on road is as given below:

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25g of mass will require 25g of opposite force on the ball from the road and opposition is moving upward to work on the ball

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3 years ago

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3 years ago

A particular conductor has 3.0 × 10^27 mobile electrons per cubic meter. The material is in the shape of a cylinder of length 6.

irina1246 [14]

Answer:

Explanation:

Given

Length of cylinder $L=6 cm$

diameter of cylinder $d=1 cm$

Area of cross-section $A=\frac{\pi\times d^2}{4}$

$A=\frac{\pi \times 10^{-4}}{4}=7.855\times 10^{-5} m^2$

$Emf =2.5 V$

current $I=5 mA$

$Resistance=\frac{V}{I}=\frac{2.5}{5\times 10^{-3}}$

$R=500 \Omega$

R is also given by

$R=\rho \frac{L}{A}$

where $\rho =resistivity$

$\rho =\frac{RA}{L}=\frac{500\times 7.85\times 10^{-3}}{6}$

$\rho =0.654 \Omega -m$

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3 years ago

Read 2 more answers

A spring of negligible mass has force constant of 1600 Newtons per meter. (a) How far must the spring be compressed for 3.20 Jou

frez [133]

Answer:

a) x=63.0 $x10^{-3} m$ or 6.3cm

b) x=116.0 $x10^{-3} m$ or 11.6cm

Explanation:

a).

The elastic potential energy is modeling by equation :

$U1=\frac{1}{2}*k*x^{2} \\K=1600 \frac{N}{m}\\ U=3.2J\\m=1.2kg\\x^{2}=\frac{2*U}{k}\\x=\sqrt{\frac{2*3.2J}{1600 \frac{N}{m}}} \\x=\sqrt{4x10^{-3} m^{2}}\\ x=0.06324m$

b).

The work energy theorem explain which work is done in this case. the motion began from the rest so K1=K2 equal zero, Ug1 is no yet done and U2is also zero because is the potential energy

$Ug=K2\\mgy=\frac{1}{2}*k*x^{2} \\but \\y=h+x=0.80m+x\\m*g*(0.80m+x)=\frac{1}{2}*k*x^{2}$

Solving for x

$2*(m*g(h+x))=k*x^{2} \\k*x^{2}-(2*m*g*x)-(2*m*g*h)=0\\1600x^{2} -2*1.2kg*9.8\frac{m}{s^{2}}*x-2*1.2kg*0.80m=0\\1600x^{2} -23.52x-18.816m=0$

$x=-\frac{b+/-\sqrt{b^{2}-4*a*c } }{2*a} \\x=-\frac{-23.52+/-\sqrt{-23.52^{2}-4*1600*-18.816 } }{2*a} \\x=11.79 +/- 173.9\\x1=-0.10 m\\x2=0.116$

The negative is discard so

x=0.116m

7 0

3 years ago