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BabaBlast [244]
2 years ago
7

Need help ASAP Thankss + BRAINLIST only for correct answer

Physics
2 answers:
spayn [35]2 years ago
7 0

Answer:

False

true

Hope this helps!

grigory [225]2 years ago
7 0
False true (person on top is correct)
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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.

3 0
3 years ago
Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th
kkurt [141]

<em></em>

Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

<em>1. Because the change in position delta X is zero.</em>

<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>

8 0
3 years ago
A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr
jasenka [17]

Answer: The original temperature was

T_{1}=126.51K

Explanation:

Let's put the information in mathematical form:

V_{1}=12.7m^{3}

T_{1}=?

V_{2}=32.5m^{3}

T_{2}=323K

P_{1}=P_{2}=3atm

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

PV=nRT

were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)

From this, taking R=0.08205746\frac{atm.l}{mol.K},  we have:

n=\frac{P_{2}V_{2}}{RT_{2}}

⇒n=3.67mol

Now:

T_{1}=\frac{P_{1}V_{1}}{nR}

⇒T_{1}=126.51K

7 0
2 years ago
Read 2 more answers
[BWS.02]If the same experiment is repeated in different parts of the world by different scientists,
guajiro [1.7K]

Answer:

the results will be the same.it may be

7 0
3 years ago
(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en
V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0
2 years ago
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