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aleksley [76]
3 years ago
8

Pls help me to draw it . I will make you brianliest ​

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
5 0
I can’t do that you have to!
You might be interested in
A cyclist maintains a constant velocity of
tensa zangetsu [6.8K]

Hi there!

We can use the equation:

d = x₀ + vt, where:

x₀ = initial distance from the reference point

v = velocity (m/s)

t = time (sec)

Plug in the given values:

d = 248 + 5(49)

d = 493m

6 0
3 years ago
A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 4.79 g coins stacked over the 28.8 cm mark, th
masha68 [24]

To solve the exercise, the key concept to be addressed is the Mass Center.

The center of mass of an object is measured as,

X_{cm} = \frac{\sum m_ix_i}{\sum m_i}

X_{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}

Our values are given by,

x_1 = 50cm

m_1 = ?

x_2 =28.8cm

m_2 = 4.79g

x_3 = 28.8cm

m_3 = 4.79g

X_{cm} = 38.4cm

Replacing the values in our previous equation we have,

X_{cm} = \frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}

38.4 = \frac{m_1(50)+2(28.8*4.79)}{m_1+2*4.79}

38.4(m_1+2*4.79)= m_1(50)+2(28.8*4.79)

38.4m_1 +367.872 = 50m_1+275.904

11.6m_1 = 91.968

m_1 = 7.928g

Therefore the mass of the meter stick is 7.928g

7 0
3 years ago
What is the frequency of a wave travelling at 45.45 m/s with a wavelength of 2 m?
HACTEHA [7]

Answer:

22.73 Hz

Explanation:

Frequency = 45.45/2 = 22.73 Hz

4 0
3 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
3 years ago
If a car accelerates uniformly from rest to 15 meters
Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0
3 years ago
Read 2 more answers
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