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4 years ago

PLEASE HELP ME I HAVE 2 MINUTES ON THIS! DO NOT GUESS!

Physics

2 answers:

11111nata11111 [884]4 years ago

4 0

Flat on her back, her whole body would be against the floor and she wouldn't have to do anything to hold herself up.

Murljashka [212]4 years ago

3 0

Yeah, she's right. Because in all the other positions, your muscles are working to keep yourself upright. But when you are laying down, you are using no muscles so you will cool off faster.

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What is environmental cleanliness

maw [93]

Answer:

it's the condition of the Earth to become neat and organized. it will give negative impact on the society and teach people to clean to have a better world to stay

8 0

3 years ago

What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields a

andrezito [222]

Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

6 0

4 years ago

You ride your bike from home to school, covering a distance of 3 km in 15 min. what is your average speed in km/h?

MrRa [10]

15 minutes = .25 hrs
3km/.25 hrs

3 0

3 years ago

A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin

gogolik [260]

Answer:

a) 23 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum p₀, can be written as follows:

$p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

The final momentum pf, can be written as follows:

$p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

$v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.

6 0

3 years ago

what is the average velocity if the initial velocity of an object is 10 m/s and the final velocity is 28 m/s?

sashaice [31]

19 m/s is the average velocity if the initial velocity of an object is 10 m/s and the final velocity is 28 m/s.

<u>Explanation:</u>

The initial and final velocity is given and the average velocity is asked, which can be calculated by the following method,

Given that initial velocity = 10 m/s and final velocity = 28 m/s

As we know that the average velocity = $\frac{\text {final velocity}+\text {inital velocity}}{2}$

Although, $\text {averge velocity}=\frac{\text {total displacement}}{\text {total time}}$

But here, there is only given initial and final velocity therefore it will be simply an average of the two.

$\text {average velocity}=\frac{10+28}{2}=\frac{38}{2}=19 \mathrm{m} / \mathrm{s}$

Therefore, the average velocity of the object is 19 m/s.

3 0

3 years ago

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