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2 years ago

A baseball is hit high and far. Which of the following statements is true? At the highest point,

Physics

1 answer:

erma4kov [3.2K]2 years ago

6 0

Answer:

option C

Explanation:

When the baseball is hit high and far it has some horizontal velocity at the highest point so option an i.e. velocity at the highest point is eliminated.

the baseball act as a projectile at the highest point so, there is always downward acceleration acting on the body. so, option e and d is eliminated.

At the highest point, the vertical velocity is zero and neglecting the air resistance we can say that the velocity of the body is slowest.

Hence, the correct answer is option C

You might be interested in

A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp

KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

$E=U=\frac{1}{2}kA^2$ (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0

2 years ago

A rectangular beam 10 cm wide, is subjected to a maximum shear force of 50000 N, the corresponding maximum shear stress being 3

nika2105 [10]

Answer:

Option B is the correct answer.

Explanation:

Shear stress is the ratio of shear force to area.

We have

Shear stress = 3 N/mm² = 3 x 10⁶ N/m²

Area = Area of rectangle = 10 x 10⁻² x d = 0.1d

Shear force = 50000 N

Substituting

$\texttt{Shear stress}=\frac{\texttt{Shear force}}{\texttt{Area}}\\\\3\times 10^6=\frac{50000}{0.1d}\\\\d=0.1667m=16.67cm$

Width of beam = 16.67 cm

Option B is the correct answer.

6 0

2 years ago

A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7

maxonik [38]

First, let's find the speed $v_i$ of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
$p_i = m_1 v_1$
After the collision, the two blocks stick together and so now they have mass $m_1 +m_2$ and they are moving with speed $v_i$ :
$p_f = (m_1 + m_2)v_i$
For conservation of momentum
$p_i=p_f$
So we can write
$m_1 v_1 = (m_1 +m_2)v_i$
From which we find
$v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s$

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force $\mu (m_1+m_2)g$ . The work done by the frictional force to stop the two blocks is
$\mu (m_1+m_2)g d$
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
$\frac{1}{2} (m_1+m_2)v_i^2$
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
$\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d$
From which we can find the value of the coefficient of kinetic friction:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

3 0

2 years ago

Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s

USPshnik [31]

No because an atom consists of two main parts and three subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.

7 0

2 years ago

Read 2 more answers

You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p

nydimaria [60]

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

X = √ (R² + ( $X_{C}$ - $X_{L}$ )² )

where Xc and XL is the capacitive and inductive impedance, respectively

X_{C} = 1 / wC

X_{L} = wL

From this expression we can see that for the resonance frequency

X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

V = IR

Since the contribution of the two other components is canceled, this occurs for

X_{C} = X_{L}

1 / wC = w L

w = √ 1 / CL

6 0

2 years ago