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3 years ago

Does anyone know how to do this? i tried doing some but i don't think i did them right.

Physics

1 answer:

geniusboy [140]3 years ago

6 0

When the velocity of an object is constant, there is no acceleration . Acceleration is defined as a change in velocity over time.
Problem three wants to know if velocity and acceleration have the same signs. I'll assume that on the x-axis the kick goes to the right, that means that velocity has a positive direction therefore positive, acceleration however, is slowing down therefore negative, this is commonly mistaken as deceleration but it's just negative acceleration.
4. when there's constant acceleration, velocity is changing all the time
(constant acceleration would be seen as a parabolic motion where velocity is seen changing at every point.)
you can use the four kinematic equations to find several things like time, distance, and acceleration for the rest

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T.E = K.E + P.E

Rasek [7]

Answer:

T.E=K.E+P.E

:146,000J=81250J+K.E

4 0

2 years ago

HNO3 (aq) + H20 (1) ► NO3- (aq) + H30+ (aq)

vodomira [7]

Answer:

It's H3O+.

Explanation:

The H3O+ ion in the solution is what gives it its acidic properties ( sour taste, low pH , for example). Its known as the hydronium ion.

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3 years ago

A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts

nalin [4]

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

$F_{x} = F - F_{air}$ (3)

Forces in the y axis:

$F_{y} = 0$ (4)

Solving for the forces in the x axis:

$F_{x} = F - F_{air}$

Where $F = 1.150x10^{3} N$ and $F_{air} = 795 N$ :

$F_{x} = 1.150x10^{3} N - 795 N$

$F_{x} = 355 N$

Replacing in equation (2) it is gotten:

$Fx + Fy = ma$

$355 N + 0 N = (90.0 Kg)a$

$355 N = (90.0 Kg)a$

$a = \frac{355 N}{90.0Kg}$

$a = \frac{355 Kg.m/s^{2}}{90.0Kg}$

$a = 3.94 m/s^{2}$

So the acceleration for the cyclist is $3.94 m/s^{2}$ , now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since he was originally at rest its initial velocity will be zero ( $v_{i} = 0$ ).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}$

$v_{f} = 16.60m/s$

He has a speed of 16.60m/s after 35.0 meters

8 0

3 years ago

Suppose that the distance an aircraft travels along a runway before takeoff is given by Upper D equals (5 divided by 3 )t square

Dahasolnce [82]

Answer:

a. Time=25seconds

b.distance=1041.67m

Explanation:

a.The equation for $D$ in terms of m/s is $\frac{250}{3}m/s$ after conversion.

To find when speed reaches 300km/hr=83.33m/s, we find $D\prime$ and solve for $t$

$D=\frac{5}{3}t^2\\D\prime=\frac{5}{3}(2t)=\frac{10}{3}t=\frac{250}{3}\\t=25sec$

b. From a, above we already have our t=25seconds as the time it takes before the plane is airborne.

#To find distance travelled in that time , we substitute for $t=25$ in our distance equation:

$D(25)=\frac{5}{3}(25)^2\\=1041.67m$

Hence the distance of the plane before it gets airborne is 1041.67m

4 0

3 years ago

the gravitational force exerted on a baseball is 2.21 N down, a pitcher throws the ball horizontally with velocity 18.0 m/s by u

Ilia_Sergeevich [38]

To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. 
b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. 
f = (ma), = .226 x 105.88, = 23.92N.

8 0

3 years ago

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