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hichkok12 [17]
3 years ago
12

An object falls from a high building and hits the ground in 8.0 seconds. Ignoring air resistance, what is the distance that it f

ell?
Physics
1 answer:
d1i1m1o1n [39]3 years ago
4 0

Answer:

310 meters

Explanation:

Given:

v₀ = 0 m/s

t = 8.0 s

a = -9.8 m/s²

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (8.0 s) + ½ (-9.8 m/s²) (8.0 s)²

Δy = -313.6

Rounded to two significant figures, the object fell 310 meters.

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21. A fisherman catches two sturgeons. The smaller of the two has a measured length of 93.46 cm (two dec- imal places and four s
Leto [7]

Answer:

135.3 cm + 93.46cm

= 228.76 cm

So total length of two fish is 228.76

5 0
2 years ago
A head-on, elastic collision between two particles with equal initial speed v leaves the more massive particle (mass m1) at rest
ZanzabumX [31]
<span>1/3 The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r" The equation for kinetic energy is E = 1/2MV^2. So the energy for the system prior to collision is 0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5 The energy after the collision is 0.5rv^2 Setting the two equations equal to each other 0.5r + 0.5 = 0.5rv^2 r + 1 = rv^2 (r + 1)/r = v^2 sqrt((r + 1)/r) = v The momentum prior to collision is -1r + 1 Momentum after collision is rv Setting the equations equal to each other rv = -1r + 1 rv +1r = 1 r(v+1) = 1 Now we have 2 equations with 2 unknowns. sqrt((r + 1)/r) = v r(v+1) = 1 Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r. r(sqrt((r + 1)/r)+1) = 1 r*sqrt((r + 1)/r) + r = 1 r*sqrt(1+1/r) + r = 1 r*sqrt(1+1/r) = 1 - r r^2*(1+1/r) = 1 - 2r + r^2 r^2 + r = 1 - 2r + r^2 r = 1 - 2r 3r = 1 r = 1/3 So the less massive particle is 1/3 the mass of the more massive particle.</span>
8 0
3 years ago
Read 2 more answers
Because atoms of elements in the same group of the periodic table have the same number of neutrons, they have similar properties
AURORKA [14]
False. What actually determines the properties of elements are the electrons, or aka valence electrons. They are used to bond, which determines its properties. 
4 0
3 years ago
Dimension equation of work
kkurt [141]

Answer:

Explanation:

Work

Other units Foot-pound, Erg

In SI base units 1 kg⋅m2⋅s−2

Derivations from other quantities W = F ⋅ s W = τ θ

Dimension M L2 T−2

Idk if this is what u are looking for but i hope this help.:)

3 0
3 years ago
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
3 years ago
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