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laila [671]
2 years ago
15

Ruth is 1.23m tall lee is 6 cm shorter than ruth. how tall is lee

Physics
1 answer:
charle [14.2K]2 years ago
7 0
Ok so 1.23m is 123cm, so now all you do is do 123-6=117

i hope this helped

please mark me as brainlest
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a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':
Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

v^{2}=u^{2}+2as

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

v=\sqrt{u^{2}+2as}

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s

Also, v=u+at and making t the subject of the formula

t=\frac {v-u}{a}

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s

Therefore, it needs 7.6 seconds to travel

7 0
3 years ago
Jamie is a hairstylist who works in a salon. He noticed that when many of the stylist or blow drying hair the power goes out wha
Mashcka [7]
The people are using a lot of electricity blow drying to many peoples hair so i would make a schedule so it dosent get to busy with costumers
6 0
3 years ago
Read 2 more answers
Which best describes the current atomic theory?
ivolga24 [154]
Choice-C is a correct statement.
6 0
2 years ago
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A penny is dropped from the Statue of Liberty
Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

vf=vi+at,

vf =unknown velocity  vi= initial velocity vi=0m/s  a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

vf=0m/s+(9.8m/s^2)*0.30seconds

Step 2: Solve

vf=2.94m/s

b)

Use the equation below to solve for the displacement at 0.30 seconds

x=vit+\frac{1}{2} at^{2}

Step 1: Substitute the same variables with the knowns

x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

5 0
2 years ago
A 2.8-kg cart is rolling along a frictionless, horizontal track towards a 1.2-kg cart that is held initially at rest. The carts
agasfer [191]

Answer with Explanation:

We are given that

Mass of one cart,m_1=2.8 kg

Mass of second cart,m_2=1.2 kg

Initial velocity of one cart,u_1=4.6m/s

Initial velocity of second cart,u_2=-2.7 m/s

a.Total momentum,P=m_1u_1+m_2u_2=2.8(4.6)+1.2(-2.7)

P=9.64 kgm/s

b.Velocity of second cart,v_2=0

According to law of conservation of momentum

Initial momentum=Final momentum

9.64=2.8v_1+1.2\times 0

v_1=\frac{9.64}{2.8}

v_1=3.44m/s

8 0
3 years ago
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