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2 years ago

What shape is the graph produced by a force vs acceleration graph?

Physics

1 answer:

hammer [34]2 years ago

6 0

The answer to the question<u> What shape is the graph produced by a force vs acceleration graph</u> is A. Linear

Since Force, F = ma where m = mass and a = acceleration. For constant mass, F ∝ a. That is, F is directly proportional to acceleration, a.

Since this is a linear relationship, the graph of force vs acceleration will be linear.

The answer to the question<u> What shape is the graph produced by a force vs acceleration graph</u> is A. Linear

Learn more about graphs here:

brainly.com/question/24322515

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A high jumper jumps over a bar that is 2 m above the mat. With what velocity does the jumper strike the mat in the landing area?

docker41 [41]

Answer:

The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.

Explanation:

It is given that,

A high jumper jumps over a bar that is 2 m above the mat, h = 2 m

We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

$v=\sqrt{2gh}$

g is acceleration due to gravity

$v=\sqrt{2\times 9.81\ m/s^2\times 2\ m}$

v = 6.26 m/s

So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.

8 0

3 years ago

Can the resistors in an "unbalanced" Wheatstone bridge circuit be treated as a combination of series and/or parallel resistors?

OlgaM077 [116]

Answer:

Explanation:

The resistors in a unbalanced wheat stone bridge cannot be treated as a combination of series and parallel combination of resistors.

In case of balanced wheat stone bridge, the resistors can be treated as the combination of series and parallel combination.

Here, In the balanced wheat stone bridge

R1 and R2 be in series and Ra and Rx is series and then their combination is in parallel combination.

4 0

3 years ago

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e

Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity $u=0$

acceleration $a=12.8 m/s^2$

velocity acquired by sled in $t_1$ time

$v=0+at$

$v=12.8t_1$

distance traveled by sled in $t_1 s$

$v^2-u^2=2as$

$(12.8t_1)^2-0=2\times 12.8\times s_1$

$s_1=6.4\cdot t_1^2$

distance traveled in $t_2$ time with velocity $v=12.8t_1$

$s_2=v\times t_2$

$s_2=12.8\times t_1\times t_2$

$s_2=12.8\cdot t_1\cdot t_2$

$s_1+s_2=5.37\times 10^3$

$6.4t_1^2+12.8t_1t_2=5370$ ----1

$t_1+t_2=97.7 s$

$t_2=97.7-t_1$

substitute the value of $t_2$ in 1

we get

$6.4t_1^2-1250.56t_1+5370=0$

thus $t_1=\frac{1250.56-1194.33}{12.8}=4.39 s$

$t_1=4.39 s$

5 0

3 years ago

al aplicar una fuerza de 2 N sobre un muelle este se alarga 4cm.¿cuanto se alargara si la fuerza es el triple?¿que fuerza tendri

3241004551 [841]

1) 12 cm

2) 3 N

Explanation:

The relationship between force and elongation in a spring is given by Hooke's law:

$F=kx$

where

F is the force applied

k is the spring constant

x is the elongation

For the spring in this problem, at the beginning we have:

$F=2 N$

$x=4 cm$

So the spring constant is

$k=\frac{F}{x}=\frac{2N}{4 cm}=0.5 N/cm$

Later, the force is tripled, so the new force is

$F'=3F=3(2)=6 N$

Therefore, the new elongation is

$x'=\frac{F'}{k}=\frac{6}{0.5}=12 cm$

In this second problem, we know that the elongation of the spring now is

$x=6 cm$

From part a), we know that the spring constant is

$k=0.5 N/cm$

Therefore, we can use the following equation to find the force:

$F=kx$

And substituting k and x, we find:

$F=(0.5)(6)=3 N$

So, the force to produce an elongation of 6 cm must be 3 N.

6 0

3 years ago

A 3.0 m tall, 40 cm diameter concrete column supports a 235,000 kg load. by how much is the column compressed? assume young's mo

statuscvo [17]

The Young modulus E is given by:
$E= \frac{F L_0}{A \Delta L}$
where
F is the force applied
A is the cross-sectional area perpendicular to the force applied
$L_0$ is the initial length of the object
$\Delta L$ is the increase (or decrease) in length of the object.

In our problem, $L_0 = 3.0 m$ is the initial length of the column, $E=3.0 \cdot 10^{10}N/m^2$ is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
$r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m$
and the cross-sectional area is
$A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2$
The force applied to the column is the weight of the load:
$W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N$

Now we have everything to calculate the compression of the column:
$\Delta L = \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m$
So, the column compresses by 1.83 millimeters.

3 0

3 years ago

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