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Aleksandr [31]
2 years ago
13

A ball is thrown straight upward with a velocity of 24m/s How much time passes before the ball strikes the ground? Disregard air

resistance.
Physics
1 answer:
zlopas [31]2 years ago
8 0
Knowing that when you throw something in the air, the speed will be 0m/s.
The acceleration due to gravity is 9.8m/s^2.

v2=0
v1=24
a=9.8

The formula is:
v2=v1+a(t)
v2-v1/a=t
0-24/9.8
24m/s/9.8m/s^2=t (You can choose which direction is positive and negative)
2.45s=t
t=2.45*2s (Since you calculate both going up and down)
t=4.9s


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The gravitational force between two objects has a magnitude of F. If both masses were doubled and the distance between them doub
Fofino [41]

Answer:

F' = F

Explanation:

The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:

F = \frac{Gm_1m_2}{r^2}

where,

F = Force of attraction

G = Universal gravitational costant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects

Now, if the masses and the distance between them is doubled:

F' = \frac{G(2m_1)(2m_2)}{(2r)^2}\\\\F' = \frac{Gm_1m_2}{r^2}

<u>F' = F</u>

7 0
2 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

5 0
3 years ago
Please help ASAP.
Bess [88]

Im not 100% sure you have to tell me if im wrong or not.

D

B

C

3 0
3 years ago
Read 2 more answers
A water pump pumps 3 kg of water every second from the bottom of a well to the top of the well 50 m above if the pump used 2000
vesna_86 [32]

watts = work per second.

work is mgh = 3x10x50=1500

watts out = 1500

Watts used = 2000

eff=1500/2000=75%


8 0
3 years ago
A rocket travels at a speed of 14,000 m/s. What is the distance travelled by the rocket in 150s?
kati45 [8]

1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

v=\frac{S}{t}

Rearranging the equation, we can write

S=vt

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m


2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s

5 0
3 years ago
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