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Irina18 [472]
3 years ago
12

If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity

would be
most useful?

a. the vehicle's speed

b. the vehicle's distance

c. the vehicle's direction

d. the vehicle's displacement
Physics
1 answer:
Papessa [141]3 years ago
8 0

Explanation:

the vehicles displacement, since displacement deals with position

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What is the matching nitrogen base sequence for the gene below?
grigory [225]

Answer:

A

Explanation:

T-G-T is the answer .

the matching Nitrogen base sequence for gene ATC is A

6 0
3 years ago
I drop a meter stick , and catch it as it falls. If it fell exactly 22 cm, how much did it take me to catch the meter stick (wha
statuscvo [17]
D=Vot+1/2at^2

In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2

acceleration=acceleration due to gravity=-9.8m/s^2

It falls - 22cm or -0.22m

We have - 0.22=1/2(-9.8)t^2

t^2=(-0.44)/(-9.8)

t=sqrt[0.44/9.8]
3 0
3 years ago
A certain superconducting magnet in the form of a solenoid of length 0.24 m can generate a magnetic field of 7.0 T in its core w
tatiyna

Answer:

N=14854.5turns

Explanation:

Given data

Length L=0.24 m

Magnetic field B=7.0 T

Current I=90 A

To find

Number of turns N

Solution

As we know that magnetic field is given as:

B=u_{o}I\frac{N}{L}

Rearrange the equation and solve for N

So

B=u_{o}I\frac{N}{L}\\\frac{B}{u_{o}I} =\frac{N}{L} \\N=\frac{LB}{u_{o}I}\\N=\frac{(0.24m)(7.0T)}{4\pi *10^{-7}*(90A)} \\N=14854.5turns

5 0
3 years ago
PLS HELP, WILL GIVE BRAINLIEST!!!
Yuki888 [10]

Answer:

b bro it's b bro

6 0
2 years ago
When a hot object is placed in a water bath whose temperature is 25◦C, it cools from 100 to 50◦C in 150 s. In another bath, the
mixas84 [53]
Using Newton's Law of Cooling, the given are:
Temperature of the first bath (Ta1): 25C
a(final temperature)= 50C, b (initial temperature)=100C
x-y is the difference in time. In the first bath, 150s

dT/dt=-k(T-Ta)
\int\limits^a_ b{(dT/(T-Ta1))} \, dt = \int\limits^x_y{-k} \, dt
ln (T-25) [from a=100 to b=50]= -k*150
ln ((50-25)/(100-25))= -k*150

Solving for k=7.324081X10^-3

For the second bath: find Ta2 with dt=120s

dT/dt=-(7.324081x10^-3)*(T-Ta2)
\int\limits^a_ b{(dT/(T-Ta2))} \, dt = \int\limits^x_y{-7.324081x10^-3} \, dt
ln (T-Ta2) [from a=100 to b=50]= -(7.324081x10^-3)*120
ln ((50-Ta2)/(100-Ta2))= -(<span>7.324081x10^-3)</span>*120

Solving for Ta (second bath temperature)= 14.49 degrees Celsius

6 0
3 years ago
Read 2 more answers
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