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olga nikolaevna [1]
2 years ago
15

A sample from of water is heated with 5000 J of energy and its temperature goes up by 6 K. What is the mass of the sample?

Physics
1 answer:
Dima020 [189]2 years ago
8 0

Answer:

mass= 0.1993 kg

Explanation:

Using the formula c = Q / (mΔT)

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Marie is puzzled by her findings she has done several meticulous calculations and has gotten the numbers .37 rad, .89 rad and 1.
Irina-Kira [14]

Answer:

I think the answer is a

Explanation:

for it to be accurate has be to exactly 0.9 rad

it is not precise because the answer she is getting is different everytime and not even close. For instance,

It would have been precise if she had gotten 0.37 rad in every attempt. or 0.89 every attempt...

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Spider-Man and Ned were testing the distance he could shoot his web depending on the angle at which he points his web shooter. H
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A dog running to the right at 4 m/s sees a ball and accelerates steadily to catch it. The dog accelerates to the right at a rate
antoniya [11.8K]

Answer:

D.-4.798m/s

Explanation:

Greetings !

Given values

u= 4ms \\ a = 0.21ms {}^{2}  \\ t = 3.8sec

Solve for V of the given expression

Firstly, recall the velocity-time equation

v = u + at

plug in known values to the equation

v = (4) + (0.21)(3.8)

solve for final velocity

v = 4.792ms

Hope it helps!

6 0
1 year ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

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3 years ago
A 12.0-g bullet is fired horizontally into a 112-g wooden block that is initially at rest on a frictionless horizontal surface a
sergiy2304 [10]
The speed would be in a decimal? Or do you want it in a fraction?
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