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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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The answer in Meters is going to to 1265.341
7 0
3 years ago
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Part A
geniusboy [140]

Answer:

a) b = -5

b) slope = 3/2

Explanation:

a) The equation of a line is given as y = mx + b, where m is the slope of the line and b is the intercept on the y axis.

Given that y = 3x + b and it passes through the point (2, 1). Hence when x = 2, y = 1. Therefore, substituting for x and y:

1 = 3(2) + b

1 = 6 + b

b = 1 - 6

b = -5

b) The equation of a line passing through two points (x_1,y_1) and x_2,y_2 is given by:

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

The equation of the line passing through the two points (0,3) and (4,9) is:

y-3=\frac{9-3}{4-0}(x-0)\\ \\y-3=\frac{3}{2}x\\ \\y = \frac{3}{2}x+3

Comparing y = (3/2)x + 3 with y = mx + b, the slope (m) is 3/2

4 0
2 years ago
A child is riding in a wagon. What reference frame might have been used if an observer said the child was not moving.
GarryVolchara [31]

Answer:

the wagon should be used as frame of reference if an observer said the child was not moving.

Explanation:

The state of motion of a body depends upon the frame of reference. It is the set of co-ordinates according to which the motion is analyzed. If a child is riding in a wagon, then he will be considered in motion to a person standing outside the wagon. Hence, if we take a frame of reference outside the wagon then the child must be in motion with respect to the observer. On the other hand if the observer is inside the wagon, then the child must be in rest with respect to the observer. Hence, if we take the wagon to be the frame of reference, then the child will be at rest with respect to the observer.

<u>Therefore, the wagon should be used as frame of reference if an observer said the child was not moving.</u>

3 0
3 years ago
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes th
Sonbull [250]

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 329.21²+39.63²

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V² = 109,949.761

V = √ 109,949.761

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Hence the speed of the shell as it hits the ground is 331.59m/s

7 0
2 years ago
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Answer:

a = 2.22 [m/s^2]

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We have to use the following kinematics equation:

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where:

Vf = final velocity = 11.11 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 5 [s]

The initial speed is taken as zero, as the car starts from zero.

11.11 = 0 + (a*5)

a = 2.22 [m/s^2]

6 0
2 years ago
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