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kap26 [50]
4 years ago
11

Which of the following statements is correct?

Physics
1 answer:
Dafna11 [192]4 years ago
6 0
These are the correct solutions:

It is 11 a.m. in the Eastern Time Zone; therefore, it is 8 a.m. in the Pacific Time Zone. (3 hrs behind)

It is 3 p.m. in the Central Time Zone; therefore, 2 p.m. in the Mountain Time Zone. (1 hr behind)

It is 6 p.m. in the Pacific Time Zone; therefore, it is 4 p.m in Hawaii. (2 or 3 hours behind depending on time of year)

It is 6 p.m. in Hawaii; therefore, it is 11 p.m. in the Eastern Time Zone (5 or 6 hours behind depending on time of year).

It is 3 p.m. in Hawaii; therefore, it is 6 p.m. in the Mountain Time Zone (3 or 4 hours behind depending on time of year).
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A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the
ivolga24 [154]

Answer:

The acceleration of the wallet is 3\hat{i}+6\hat{j}

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse a=2\hat{i}+4.00\hat{j}

We need to calculate the acceleration of the wallet

Using formula of acceleration

a=r\omega^2

Both the purse and wallet have same angular velocity

\omega=\omega'

\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}

\dfrac{a}{r}=\dfrac{a'}{r'}

\dfrac{a'}{a}=\dfrac{r'}{r}

\dfrac{a'}{a}=\dfrac{3.45}{2.30}

\dfrac{a'}{a}=\dfrac{3}{2}

a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})

a'=3\hat{i}+6\hat{j}

Hence, The acceleration of the wallet is 3\hat{i}+6\hat{j}

4 0
3 years ago
A greyhound's velocity changes from 10 meters per second to 15 meters per second in 0.5 second. What is the greyhound's average
Elden [556K]

The answer would be 10ms 2

4 0
4 years ago
CAN SOMEONE HELP ME PLEASE!? After chasing its prey, a cougar leaves skid marks that are 236 m in length. Assuming the cougar sk
malfutka [58]

Answer:

u=36.8m/s

Explanation:

because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations

u^2=v^2-2ā*s. where:

u^2 stands for intial velocity

v^2 stands for final velocity

since the cougar skidded to a complete stop the final velocity is zero.

u^2=v^2-2ā*s

u^2=(0)^2 -2(-2.87 m/s^2)*236 m

u^2=0+5.74m/s^2* 236m

u^2=1354.64m^2/s^2

u=√1354.64m^2/s^2

u=36.8m/s (approximate value)

when ever the acceleration is constant you can use one of the following equation to find the required value.

1. v = u + at. (no s)

2. s= 1/2(u+v)t. (no ā)

3. s=ut + 1/2at^2. ( no v)

4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)

5 0
3 years ago
a fast humvee drove from desert a to desert b. for the 12 hours, it travelled at an speed of 185 km/h. for the next 13 hours, it
ivanzaharov [21]

Answer:

I am not sure of the question?

Explanation:

List the question

8 0
3 years ago
1. Why didn't Sucheng Chan go to school until she was eight
Crazy boy [7]

Answer:

Because china was in war, and her parents didn't have enough money for to go to school.

3 0
3 years ago
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