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Ray Of Light [21]
3 years ago
8

A ball with an initial velocity of 25 m/s is subject to an acceleration of -9.8m/s^2 how high does it go before coming to a mome

ntary stop ?

Physics
1 answer:
qwelly [4]3 years ago
8 0

Answer:

The ball will reach an height of 31.88 meters before coming to a stop.

Explanation:

we know that by equation of motion v^{2} = u^{2} + 2as

where v is the final velocity of the ball , u is the initial velocity of the ball , a is acceleration due to gravity and s is the distance traveled by the ball.

a = -9.8 \frac{m}{s^{2}}

and u = 25 m/s

and v = 0

therefore s = \frac{u^{2}}{2a}

so s = 31.88 meters

You might be interested in
Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m
Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, t_t of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + t_t + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0
3 years ago
Multiply 5.036×102m by 0.078×10−1, taking into account significant figures.
Anika [276]

Answer : The significant digit is 6

Explanation :

Multiply 5.036\times10^{2}m by 0.078\times10^{-1}m

Now, on multiplying

5.036\times10^{2}m \times0.078\times10^{-1}m = 0.392808 \times10^{1}\ m^{2}

5.036\times10^{2}m \times0.078\times10^{-1}m = 0.0392808\ m^{2}

Now, the significant digit is 6.

Hence, this is the required solution.

3 0
3 years ago
Why is dose equvialent in rem or sv used to define the Effective Dose Equivalent and the Committed Dose Equivalent?
Komok [63]

Good question next question

8 0
3 years ago
calculate the spring constant if a weight of 250N is added to a spring which increases in length by 20cm
ZanzabumX [31]
Since, F = k . ∆x

Therefore, k = F / ∆x = 250 / 0.2 = 1250 N/m

(ps: convert 20 cm into 0.2 m)
8 0
3 years ago
an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and
gavmur [86]

Answer:

Density of Sand is 2.653g/cm^{3}.

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle(w_{max})=48.4gm-23.5gm

w_{max}=24.9g

Since we know density of water d_{w} =1g/cm^{3} and w_{max}

We can calculate volume of empty space in the bottle(V).

w_{max}=d_{w}\timesV

V=\frac{w_{max} }{d_{w} }

V=\frac{24.9}{1}

V=24.9 g/cm^{3}

Now bottle is partially filled with sand,and weight of bottle is (w_{s})36.5gm

So,

Amount of sand added (m_{s})=36.5-Weight of the bottle

m_{s}=13g

After filling the bottle with water again,the weight of the bottle becomes (W_{2}=56.5g)

Therefore,

amount of water added to the bottle of sand in grams = W_{2}-36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/cm^{3}

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle(V_{w})=20cm^{3}

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V-V_{w}

V_{s}=24.9g-20g

V_{s}=4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = \frac{m_{s} }{V_{s} }

density of sand =\frac{13}{4.9}

density of sand = 2.653g/cm^{3}

8 0
3 years ago
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