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dem82 [27]
3 years ago
5

Please check my work

Physics
2 answers:
Andrej [43]3 years ago
7 0

Answer:

All correct

Explanation:

ira [324]3 years ago
6 0
Yea some of them are not correct but some are.
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a baseball is pitched with a speed of 35.0m/s. how long does it take the ball to travel 18.4 m from the pitchers mound to home p
Vlad [161]
Given the speed and the distance, to find time you can use the formula speed is equal to distance over time. From there you can manipulate the equation for time to equal the distance divided by speed. Time is equal to 18.4 meters divided by 35m/s which equals 0.526 seconds.

7 0
3 years ago
Kelly is riding a bicycle, moves with an initial velocity of 5 m/s. Ten seconds later, she is moving at 15 m/s. What is her acce
sladkih [1.3K]

Answer:

her acceleration is 1 m/sec

Explanation:

The following information is given in the question

The initial velocity is 5 m/s

After 10 seconds, she would be moved at 15 m/s

We need to find the acceleration

As we know that

Acceleration = Change in speed ÷ time

Acceleration = (15 - 5) ÷ (10)

= 1 m/sec

Hence, her acceleration is 1 m/sec

The same would be considered  

3 0
3 years ago
Hello Peeps can y'all PLEASE HELP ME? I need to graduate!
11111nata11111 [884]
1. Accelarating
2. Constant Speed
3. Decelerating
4. At Rest
5. Accelerating
Hope this helped!
6 0
4 years ago
You wish to cook some pasta and so take some water and heat it on a stove. If you use 2000 grams of water with an initial temper
Ronch [10]

Answer:

the stove energy went into heating water is 837.2 kJ.

Explanation:

given,

mass of water = 2000 grams

initial temperature = 0° C

Final temperature = 100° C

specific heat of water (c) = 4.186 joule/gram

energy = m c Δ T                            

            = 2000 × 4.186 × (100° - 0°)

            = 837200 J                          

             = 837.2 kJ            

hence, the stove energy went into heating water is 837.2 kJ.

3 0
3 years ago
In Fig., block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.50 m and then collides with stationary
Evgesh-ka [11]

Answer:

Explanation:

Given

initial height h=2.5 m

m_2=2m_1

coefficient of static friction \mu =0.5

When collision is elastic respective velocities after collision is

v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}

v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}

where u_1 and u_2=initial velocities of object

v_1 and v_2 final velocities of object

u_1=\sqrt{2\times 9.8\times 2.5}

u_1=7 m/s

v_2=\frac{0+2m_1\times 7}{m_1+2m_1}

v_2=\frac{14}{3} m/s

using v^2-u^2=2 as

0-(4.67)^2=2\times (-0.5\times 9.8)\times s

s=2.22\ m

(b)Completely Inelastic

In Completely Inelastic objects stick with each other

m_1u_1=(m_1+m_2)v

v=\frac{u_1}{3}=\frac{7}{3} m/s

using v^2-u^2=2 as

0-(2.33)^2=2\times (-0.5\times 9.8)\times s

s=0.55\ m                          

7 0
4 years ago
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