Answer:
5.52 m
Explanation:
Using the equation;
Vy = Vo sin θ
Where; Vo = 2.25 m/s and θ = 35°
We get;
= 2.25 sin 35°
= 1.29 m/s
Thus; Vy = 1.29 m/s
But a = -9.81 m/s (against gravity)
t = 1.20 s and
y = Vyt + 1/2 at²
where y is the height above the water
= 1.29 × 1.2 + (1/2 × -9.81 × 1.2²)
= 1.548 + (-7.0632)
= - 5.5152 m
Thus; the height of the girl above the water is 5.52 m
Refer to the diagram shown below.
Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.
Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.
Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s
100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s
V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°
Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s
The circuit is in parallel connection
Equivalent resistance = 1/Req = 1/R1 + 1/R2 + 1/R3
From the information given,
R1 = 5
R2 = 2
R3 = 4
1/Req = 1/5 + 1/2 + 1/4 = (4 + 10 + 5)/20 = 19/20
Req = 20/19 = 1.053 ohms
I = V/R
Given that V = 12,
Current flow through circuit = 12/1.053 = 11.4 A
I1 + I2 + I3 = 11.4
I1 = 12/5 = 2.4 A
I2 = 12/2 = 6 A
I3 = 12/4 = 3A
F=ma
8480=26.5m
m=8480/26.5
m=320
The mass of the cart is 320kg.
Answer:
b) The star is moving away from us.
Explanation:
If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.