All categories

3 years ago

Orlat

1 answer:

8 0

The maximum force that the athlete exerts on the bag is equal to 1,500 N and in the opposite direction as the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states that action and reaction are equal and opposite.

Fa = -Fb

The force exerted by the athlete on the bag is equal to the force the bag exerted on the athlete but in opposite direction.

Thus, the maximum force that the athlete exerts on the bag is equal to 1,500 newtons and in the opposite direction as the force that the bag exerts on the athlete.

Learn more about force here: brainly.com/question/12970081

#SPJ1

You might be interested in

In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m a

matrenka [14]

Assuming you are looking for the acceleration a:

1. $m_1a = T_1 -m_1g$
2. $m_2a = m_2g - T_2$
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3. $\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}$
where $I = \frac{1}{2} mR^2$ and $a = \alpha R$ .

Combining the three equations:
$T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }$

6 0

3 years ago

How would you find the total energy stored in the

likoan [24]

Answer:

The energy of the capacitors connected in parallel is 0.27 J

Given:

C = $2.0\micro F = 2.0\times 10^{- 6} F$

C' = $4.0\micro F = 4.0\times 10^{- 6} F$

Potential difference, V = 300 V

Solution:

Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:

$C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F$

The energy of the capacitor, E is given by;

$E = \frac{1}{2}C_{eq}V^{2}$

$E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J$

6 0

3 years ago

A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an

ANTONII [103]

$a=5000\dfrac{mm}{s}=5\dfrac{m}{s}$

$f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s$

$a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}$

Hope this helps.

r3t40

5 0

3 years ago

____ is a pesticide that influenced society by increasing crop yield, but it also caused human and animal sickness.

Ilya [14]

I think your answer would be DDT, but don't hold me to it

5 0

3 years ago

A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge

zheka24 [161]

Answer:

The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

$F_{1}=\dfrac{kq_{1}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}$

$F_{1}=8\times10^{-11}j\ \ N$

We need to calculate the force on electron due to q₂

Using formula of force

$F_{2}=\dfrac{kq_{2}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}$

$F_{2}=6.93\times10^{-11}i\ \ N$

We need to calculate the net force

Using formula of net force

$F=F_{1}+F_{2}$

Put the value into the formula

$F=8\times10^{-11}j+6.93\times10^{-11}i$

The magnitude of the net force

$F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}$

$F=1.058\times10^{-10}\ N$

We need to calculate the acceleration of an electron

Using formula of force

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}$

$a=1.2\times10^{20}\ m/s^2$

Hence, The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

3 0

4 years ago