All categories

2 years ago

Grrr i don't have much time, help please T^T

Physics

2 answers:

mart [117]2 years ago

4 0

Answer:

2. ( b ) zero

3. ( c ) 10 s

4. Uniform then decreasing

Explanation:

Since the motion is uniform, initial and final velocity will be 0, hence acceleration will be zero.

Initial velocity ( u ) = 5 m/s

Final velocity ( v ) = 35 m/s

Acceleration ( a ) 3 m/s^2

To find : Time ( t )

Formula : -

t = v - u / a

= 35 - 5 / 3

= 30 / 3

t = 10 s

alekssr [168]2 years ago

4 0

Answer:

See below ~

Explanation:

Q2

The uniform circular motion in the movement of object in a circular path with fixed and perpendicular on the direction of the motion acceleration.

Q3

Substitute the values in : v = u + at
35 = 5 + 3t
3t = 30
t = 10s (c)

Q4

The vertical component for a projectile by acute angle on the horizontal axis changes from decreasing then increasing velocity.

You might be interested in

At sunrise two old women started to walk towards each other. one started from point a and went towards point b while the other s

konstantin123 [22]

Let the distance between points a and b = d

For the women starting from a her velocity be $V_1$ and for the women starting from b her velocity be $V_2$

Let the rising time of sun = x AM

We have $V_1$ = $\frac{d}{12+4-x}$

and we also have $V_2$ = $\frac{d}{12+9-x}$

At they meet means, they both travel a combined distance of d

So we have $V_1*(12-x)+V_2*(12-x)=d$

Substituting velocity values we will get

$\frac{d}{12+4-x}*(12-x)+\frac{d}{12+9-x}*(12-x)=d\\ \\ \frac{12-x}{12+4-x}+\frac{12-x}{12+9-x}=1\\ \\ \frac{12-x}{16-x}+\frac{12-x}{21-x}=1\\ \\ 252-21x-12x+x^2+192-12x-16x+x^2=336-16x-21x+x^2\\ \\ x^2-24x+108=0\\ \\ (x-6)(x-18)=0\\ \\$

x= 6 AM or 18 AM, but 18 AM is not possible,

Sun rise time is 6 AM

3 0

3 years ago

A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energ

astra-53 [7]

Answer:

Explanation:

The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation

$v_f=v_0+at$ where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...

0 = 21.5 + (-9.8)t and

-21.5 = -9.8t so

t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)

Now we will use that time to find out the max height of the object in the equation

Δx = $v_0t+\frac{1}{2}at^2$ and filling in: