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2 years ago

14

A ball is tossed up into the air, reaches its highest point and returns to its original position. which of the following is true

about the balls velocity and acceleration at its highest point?
a) it’s velocity and acceleration are both zero

b) it’s velocity is up and non zero constant and acceleration is zero

c) it’s velocity is down and non zero constant and acceleration is zero

d) it’s velocity is zero and acceleration is up and non zero constant

e) it’s velocity is zero and acceleration is down and non constant zero

1 answer:

Nikitich [7]2 years ago

5 0

Answer:

At the top v =0 and there will be a acceleration due to gravity (g) acting downward.

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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i

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Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = $\frac{28}{10}$ × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

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3 years ago

We are able to see an object when it:

xenn [34]

Answer:

C. reflects one or more visible light frequencies.

Explanation:

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4 years ago

Read 2 more answers

0.3-L glass of water at 20C is to be cooled with ice to 5C. Determine how much ice needs to be added to the water, in grams, i

abruzzese [7]

Answer:

$a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g$

Explanation:

First we need to state our assumptions:

Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice $h_i_f=333.7KJkg$

Mass of water, $m_w=\rho V =1\times0.3=0.3Kg$ .

Energy balance for the ice-water system is defined as

$E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w$

a.The mass of ice at $0\textdegree C$ is defined as:

$[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g$

b.Mass of ice at $20\textdegree C$ is defined as:

$[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g$

c.Mass of cooled water at $T_c_w=0\textdegree C$

$\bigtriangleup U_c_w+\bigtriangleup U_w=0$

$[mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g$

8 0

3 years ago

A car is traveling at the bottom of a 9.00-meter-radius circular hill with a constant speed v. The moment the car is at the bott

Rina8888 [55]

Answer:

Explanation:

reading of scale = reaction force of surface R

centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .

R = mg + m v² / R

given ,

m v² / R = .80 mg

v² = .80 x g x R

= .8 x 9.8 x 9 = 70.56

v = 8.4 m /s

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3 years ago

At the end of a delivery ramp, a skid pad exerts a constant force on a package so that the package comes to rest in a distance d

MatroZZZ [7]

Answer:

increases by a factor of $\sqrt{2}$

Explanation:

First we need to find the initial velocity for it to stop at the distance 2d using the following equation of motion:

$v^2 - v_0^2 = 2a\Delta s$

where v = 0 m/s is the final velocity of the package when it stops, $v_0$ is the initial velocity of the package when it, a is the deceleration, and $\Delta s = d$ is the distance traveled.

So the equation above can be simplified and plug in Δs = d, $v_0 = v_1$ for the 1st case

$-v_1^2 = 2ad$ (1)

For the 2nd scenario where the ramp is changed and distance becomes 2d, $v_0 = v_2$

$-v_2^2 = 4ad$ (2)

let equation (2) divided by (1) we have:

$\left(\frac{v_2}{v_1}\right)^2 = 4ad / 2ad = 2$

$v_2 / v_1 = \sqrt{2}$

$v_2 = \sqrt{2}v_1$

So the initial speed increases by $\sqrt{2}$ . If the deceleration a stays the same and time is the ratio of speed over acceleration a

$t = v / a$

The time would increase by a factor of $\sqrt{2}$

7 0

4 years ago