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lyudmila [28]
3 years ago
6

A 1200 kg vehicle initially travelling at 50. km/h experiences an air resistance of 5000 N and road friction of 2200 N. The whee

ls deliver a force of 7500 N to the vehicle. What is its acceleration?
Physics
1 answer:
True [87]3 years ago
8 0
The forces on the car are:
-- 7,500 N forward, from the engine
-- 5,000 N backwards, from air resistance
-- 2,200 N backwards, from road friction

The net force on the car is (+7,500 - 5,000 - 2,200) = 300 N forward .

Newton said: Force = (mass) x (acceleration)
Divide each side by (mass) :
Acceleration = (force) / (mass)

Acceleration = (300 N) / (1,200 kg) = <u>0.25 m/s² forward</u>.

Notice the big red herring at the beginning of the question:
It's nice to know that the vehicle starts out traveling at 50 km/hr,
but in order to answer this question, we honestly don't care !

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D. Saturn, Jupiter and Uranus
8 0
3 years ago
An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit
lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

m_2\text{ and }v_2 = Final mass and velocity

We are given:

m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s

Putting values in above equation, we get:

238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

4 0
3 years ago
An athlete needs to lose weight and decides to do it by "pumping iron." (a) How many times must an 60.0 kg weight be lifted a di
Dennis_Churaev [7]

Answer:

a) 37171

b) 20650.5 sec

Explanation:

m = mass of the weight being lifted = 60.0 kg

d = distance by which the weight is lifted = 0.670 m

E = Energy available to burn = 1 lb = 3500 kcal = 3500 x 4184 J

g = acceleration due to gravity = 9.8 m/s²

n = number of times the weight is lifted

Energy available to burn is given as

E = n m g d

3500 x 4184 = n (60) (9.8) (0.670)

n = 37171

b)

T = time period for each lift up = 1.80 s

t = total time taken

Total time taken is given as

t = \frac{n}{T}

t = \frac{37171}{1.80}

t = 20650.5 sec

8 0
3 years ago
Has anyone done this???i need the last questions for 1 and 2
Helen [10]

Answer:

no i haven't done it

Explanation:

clean yo computer tho

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A tetrameric protein dissociates into dimers when the detergent sodium dodecyl sulfate (SDS) is added to a solution of the prote
KatRina [158]

Answer:

At the dimer-dimer interface there might be acting non-covalent forces (van der waals, Hidrogene bridges, hydrophobic forces)

At the monomer-monomer interface there might be covalent forces acting (disulfide bridges).

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On the SDS-PAGE application  works by disrupting non-covalent bonds in the proteins, and so denaturing them. Therefore, the disulfide bridges won´t be disrupted, so the monomers will remain bounded.

5 0
3 years ago
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