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AleksandrR [38]
3 years ago
12

A tornado lifts a truck 252 m above the ground ñ. As the storm continues, the tornado throws the truck horizontally. It lands 56

0 m away from where it was picked up. How fast was the truck traveling horizontally through its fight? Round your answer to the nearest tenth.
Physics
2 answers:
sashaice [31]3 years ago
8 0

i think its 200 i hope that helps

Lubov Fominskaja [6]3 years ago
7 0

Answer:

78.1 m/s

Just took it.

Explanation:

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A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc
nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

m_sv_s + m_rv_r = 0

where m_s = 72.8, m_r = 265 are the mass of the swimmer and raft, respectively. v_s = 5.21 m/s, v_r are the velocities of the swimmer and the raft after the run, respectively. We can solve for v_r

265v_r + 72.8*5.21 = 0

v_b = -72.8*5.21/265 = -1.43 m/s

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0
3 years ago
A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 106.0 km/h. At that speed, the forces resistin
otez555 [7]

Answer:

"1155 N" is the appropriate solution.

Explanation:

Given:

Acceleration,

a=3 \ m/s^2

Forces resisting motion,

F_f=432 \ N

Mass,

m = 241 \ kg

By using Newton's second law, we get

⇒ F-F_f=ma

Or,

⇒         F=ma+F_f

By putting the values, we get

⇒             =(3\times 241)+432

⇒             =723+432

⇒             =1155 \ N

7 0
3 years ago
An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts
Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

7 0
3 years ago
Read 2 more answers
You are given two unknown point charges, Q1, and Q2. At a point on the line joining them, one third of the way from Q1 to Q2 the
podryga [215]
The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case, 
kQ1/(r1)^2 = kQ2/(r2)^2  r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2  kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 =   1/4
3 0
3 years ago
Read 2 more answers
A boy and his younger sister are at the zoo on a hot day. They each buy a cold lemonade. The boy buys a large lemonade, and his
FromTheMoon [43]

Answer: A and C (i took the test)

Explanation: Hope this helps:)

7 0
3 years ago
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