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3 years ago

True or false

1 answer:

8 0

That depends on what you're calling the 'x-component'.

As you increase the launch angle of a projectile, the vertical component
of its initial velocity increases, and the horizontal component of its initial
velocity decreases.

At the same time, the vertical component of its acceleration is a constant
9.8 m/s² throughout the event, while the horizontal component of its
acceleration is a constant zero.

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In boxing, the use of 16-ounce gloves rather than 12-ounce gloves reduces the chance of injury because the force is distributed

mr Goodwill [35]

Answer:

true

Explanation:

Here we have assumed that increasing the mass of a glove will increase the surface area.

Injury is caused by the application of pressure at a point on the body. The application of pressure takes place via the area of the gloves. Pressure is given by

$P=\dfrac{F}{A}$

where

F = Force

A = Area to which the force is applied

So, a bigger glove will increase the surface area and reduce the pressure resulting in a lower chance of injury.

Hence, the statement is true.

5 0

3 years ago

Pooping in my room and my room is upstairs and upstairs bathroom upstairs

diamong [38]

Answer:

huh? do you need help on math?

Explanation:

what do you mean?

7 0

2 years ago

The wave function of a particle is exp(i(kx-omegat)), where x is distance, t is time, and k and co are positive real numbers. Th

Step2247 [10]

Answer:

Momentum, $p=\hbar k$

Explanation:

The wave function of a particle is given by :

$y=exp[i(kx-\omega t)]$ ...............(1)

Where

x is the distance travelled

t is the time taken

k is the propagation constant

$\omega$ is the angular frequency

The relation between the momentum and wavelength is given by :

$p=\dfrac{h}{\lambda}$ ............(2)

From equation (1),

$k=\dfrac{2\pi}{\lambda}$

$\lambda=\dfrac{2\pi}{k}$

Use above equation in equation (2) as :

$p=\dfrac{h k}{2\pi }$

Since, $\dfrac{h}{2\pi}=\hbar$

$p=\hbar k$

So, the x-component of the momentum of the particle is $\hbar k$ . Hence, this is the required solution.

8 0

3 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at

noname [10]

$Given:\\m_b=100kg\\m_c=1600kg\\v_c=5.2 \frac{m}{s} \\p_b=p_c\\\\Find:\\v_b=?\\\\Solution:\\\\p_b=p_c\\\\p=mv\\\\m_bv_b=m_cv_c\Rightarrow v_b= \frac{m_cv_c}{m_b} \\\\v_b= \frac{1600kg\cdot5.2 \frac{m}{s} }{100kg} =83.2 \frac{m}{s}$

3 0

3 years ago

A sprinter starts from rest and runs with constant acceleration to a top speed of 12m/s in 4.0 seconds.

bagirrra123 [75]

Answer:

a = 3,0 m/s²

Explanation:

En este ejercicio se pide calcular la aceleracion del cuerpo, usemos las ecuaciones de cinematica en una dimensión.

v= v₀ + a t

como el corredor parte del reposo si velocidad inicial es cero

v = at

a = v/t

calculemos

a = 12 /4,0

a = 3,0 m/s²

4 0

3 years ago