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kozerog [31]
4 years ago
14

WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car

which accelerates from 0 to 30 m/s in 8 seconds?

Physics
1 answer:
olga nikolaevna [1]4 years ago
4 0

My Phone is +2348181686682

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1. How much heat energy ( Q ) is required to heat 2.0 kg of copper from 30.0 oC to 80.0 oC?
djverab [1.8K]

Answer:

Heat capacity, Q = 38500 Joules

Explanation:

Given the following data;

Mass = 2 kg

Initial temperature, T1 = 30°C

Final temperature, T2 = 80°C

Specific heat capacity of copper = 385 J/Kg°C

To find the quantity of heat required;

Mathematically, heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

M represents the mass of an object.

C represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 80 - 30

dt = 50°C

Substituting the values into the formula, we have;

Q = 2 * 385*50

Heat capacity, Q = 38500 Joules

4 0
3 years ago
What is the average velocity if the initial velocity of an object is 10 m/s & the final velocity is 28 m/s
Likurg_2 [28]

Answer:

v_{1}  = 19 m/s

Explanation:

v_{1}  = \frac{(v_{2}  + u)}{2}, where v_1<em> </em>= avg. velocity, v_2 = final velocity, and u = initial velocity.

6 0
3 years ago
Which best describes accuracy?
Likurg_2 [28]
The answer is either c or b
7 0
3 years ago
Read 2 more answers
A small racquet ball launcher is set up 5 meters from a 12 meter tall wall. It launches a racket ball at 30 m/s at an angle of 4
Strike441 [17]

Answer:20.82 m

Explanation:

Given

distance between wall and launcher=5 m

initial velocity=30 m/s

Launching angle=45^{\circ}

Height of wall=12 m

maximum height by ball

h_{max}=\frac{u^2sin^2\theta }{2g}

h_{max}=\frac{30^2sin^{2}45}{2\times 9.8}

h=22.95 m

y=xtan\theta -\frac{gx^2}{2u^2cos^\theta }

y=4.72 m

so ball will strike with wall

and for perfect restitution final velocity of ball will be same as the horizontal velocity before its impact, only direction will be opposite and vertical velocity will be zero

thus it seems as if someone throw the ball with horizontal velocity of 30cos45 from a height of 4.72

Time required to cover 4.72 m

t=\sqrt{\frac{2h}{g}}

t=\sqrt{\frac{9.45}{9.8}}

t=0.981 s

Horizontal distance traveled in this time

R=ucos45\times t

R=30\times cos45\times 0.981

R=20.82 m

6 0
3 years ago
An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. at time t = 0.00
Masteriza [31]
The cart is moving by simple harmonic motion, and its position at time t is described by
x(t) = A \cos (\omega t)
where
A is the amplitude of the oscillation
\omega is the angular frequency

The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released: 
A=0.250 m

The period of the motion is T=5.67 s, and the angular frequency is related to the period by
\omega =  \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s

Therefore now we can calculate the position of the system at the time t=29.6 s:
x(29.6 s)=(0.250 m)\cos ((1.11 rad/s)(29.6 s))=+0.033 m
3 0
3 years ago
Read 2 more answers
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