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kotegsom [21]
3 years ago
9

A ball thrown vertically upward is caught by the thrower after 2.93 s. Find the initial velocity of the ball. The acceleration o

f gravity is 9.8 m/s 2 . Answer in units of m/s.
Physics
1 answer:
almond37 [142]3 years ago
7 0

Answer:

The initial velocity of the ball is 28.714 m/s

Explanation:

Given;

time of flight of the ball, t = 2.93 s

acceleration due to gravity, g = 9.8 m/s²

initial velocity of the ball, u = ?

The initial velocity of the ball is given by;

v = u + (-g)t

where;

v is the final speed of the ball at the given time, = 0

g is negative because of upward motion

0 = u -gt

u = gt

u = (9.8 x 2.93)

u = 28.714 m/s

Therefore, the initial velocity of the ball is 28.714 m/s

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If a rock has a weight of 30 N on Earth, would its weight be more or less if it was on Jupiter (gravity on Jupiter = 25 m/s2)?
goldenfox [79]
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A WLAN transmitter that emites a 100 mW signal is connected to a cable with a 3dB loss. If the cable is connected to an antenna
irakobra [83]

Answer: 24 dBm

Explanation:

in the attachment

6 0
3 years ago
A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje
kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0
3 years ago
if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?
KonstantinChe [14]
Since the stone was dropped from height, its initial velocity = 0 m/s

Using  v² = u² + 2gs.

Where g ≈ 10 m/s²,  u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s 
6 0
3 years ago
Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT
Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

\\ \tt\bull\leadsto PV

\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]

\\ \tt\bull\leadsto [ML^5T^{-2}]

RHS

\\ \tt\bull\leadsto nRT

\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]

\\ \tt\bull\leadsto [ML^5T^{-2}]

LHS=RHS

hence verified

6 0
3 years ago
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