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Tpy6a [65]
3 years ago
14

Liz puts a 1 kg weight and a 10 kg on identical sleds. She then applies a 10N force to each sled. Describe why the smaller weigh

t has a larger acceleration. A) There is same amount force is applied to both sleds. B) The acceleration is directly proportional to the force. C) It indicates the extent of the unbalanced forces present. D) Acceleration depends indirectly on the mass of the object.
Physics
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0

the correct answer is D- Acceleration depends indirectly on the mass.

According to Newton's second law, the force F, the mass m and the acceleration a are related as follows:

F=ma

Therefore,

a=\frac{F}{m}

The acceleration <em>a₁ </em>of the mass<em> m₁ =1 kg</em>  is given by,

a_1=\frac{F}{m_1} \\ =\frac{10N}{1kg} \\ =10m/s^2

The acceleration <em>a₂ </em>of the mass <em>m₂=10 kg</em> is given by,

a_2=\frac{10N}{10kg} \\ =1m/s^2

The smaller mass has greater acceleration.

Thus, when the force applied on two bodies of different masses remains constant, then,

a\alpha  \frac{1}{m}

Acceleration is inversely proportional to the mass of the body.



laiz [17]3 years ago
8 0

the answer is D i had the test

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7 0
3 years ago
The drawing shows a skateboarder moving at 4.8 m/s along a horizontal section of a track that is slanted upward by 48° above the
Anna11 [10]
Let M = mass of the skier, 
v2 = his speed at the end of the track. 
By conservation of energy, 
1/2 Mv^2 = 1/2 Mv2^2 + Mgh 
Dividing by M, 
1/2 v^2 = 1/2 v2^2 + gh
 Multiplying by 2, 
v^2 = v2^2 + 2gh 
Or v2^2 = v^2 - 2gh 
Or v2^2 = 4.8^2 - 2 * 9.8 * 0.46 
Or v2^2 = 23.04 - 9.016 
Or v2^2 = 14.024 m^2/s^2-----------------------------(1) 
In projectile motion, launch speed = v2 
and launch angle theta = 48 deg 
Maximum height 
H = v2^2 sin^2(theta)/(2g) 
Substituting theta = 48 deg and value of v2^2 from (1),
 H = 14.024 * sin^2(48 deg)/(2 * 9.8) 
Or H = 14.024 * 0.7431^2/19.6 
Or H = 14.024 * 0.5523/19.6 
Or H = 0.395 m = 0.4 m after rounding off 
Ans: 0.4 m

The answer in this question is 0.4 m
4 0
2 years ago
A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula
Marina86 [1]

Answer:\alpha =10.66 rad/s^2

Explanation:

Given

mass of disk m=5 kg

diameter of disc d=30 cm

Force applied F=4 N

Now this force will Produce  a  torque of magnitude

T=F\cdot r

T=4\dot 0.15

T=0.6 N-m

And Torque is given Product of moment of inertia and angular acceleration (\alpha )

T=I\cdot \alpha

Moment of inertia for Disc I= \frac{Mr^2}{2}

I=0.05625 kg-m^2

0.6=0.05625\cdot \alpha

\alpha =10.66 rad/s^2

3 0
3 years ago
A disk with a hole has inner radius rin and outer radius rout. the disk is uniformly charged with total charge q. find an expres
Effectus [21]
Here it is.  I used the surface charge density σ as q/Area, and then wrote it out in terms of q at the end

3 0
3 years ago
How long does it take light to reach us from the Sun, 1.53x108 km away? Give your answer in minutes.
muminat

Answer:

Sun light takes 8.5 minutes to reach Earth.

Explanation:

Light travels in vacuum at constant speed c= 3 \times 10 ^8 \cfrac ms, so we can find the time it takes to travel the given distance from Earth to Sun using a kinematic equation.

Finding the time it takes for Sun light to reach us.

Using the kinematic equation d = vt, we can solve for time t, which will give us

t = \cfrac dv

We can replace the equation with the given values, but before doing that, we need the values to be on SI units, thus the distance becomes

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t = \cfrac{1.53 \times 10^{11} m}{3 \times 10^{8} \cfrac ms}\\t = 510 s

We get as result 510 seconds, then we must convert them in minutes.

t = 510 s \times \cfrac{1 min}{60s}\\\boxed{t = 8.5 min }

So we can conclude that Sun light takes 8.5 minutes to reach Earth.

7 0
2 years ago
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