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2 years ago

EASY BRAINLIEST!!URGENT PLEASE HELP.

Physics

2 answers:

zavuch27 [327]2 years ago

8 0

Answer: I believe that it is A and D. Or A or D.

A, D

Explanation:

sveta [45]2 years ago

5 0

Answer: A and C

Explanation: The rest of the thermal energy goes to the engine making it heat up thats why is can overheat also letting it get destroyed

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When iron rusts, iron atoms are destroyed true or false?

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False. this is just a physical change, not a chemical change.

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3 years ago

When you set a heavy bag down on the ground, you are doing _______ work on it.

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2 years ago

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Diano4ka-milaya [45]

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3 years ago

A simple pendulum consisting of a small object of mass m attached to a string of length l has a period T.

andrew-mc [135]

Answer:

Check Explanation.

Explanation:

For a simple pendulum, the period is given as

T = 2π√(L/g)

It is also given as

T = 2π√(m/k)

where

T = period of oscillation

m = mass of the pendulum

L = length

g = acceleration due to gravity

k = force constant

Equating this two equations,

2π√(L/g) = 2π√(m/k)

(L/g) = (m/k)

(m/L) = (k/g)

So, any pendulum that will have the same period as our pendulum with mass, m, and length, L, must have the ratio of (L/g) to be the same as the pendulum under consideration and the ratio of its mass to force constant (m/k) must also be equal to this ratio.

Hope this Helps!!!

4 0

3 years ago

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li

erica [24]

Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

Stan's speed = 42.9 m/s

Explanation:

<u>Given:</u>

$u_k$ = initial speed of Kathy = 0 m/s

$u_s$ = initial speed of Stan = 0 m/s
$a_k$ = acceleration of Kathy = $4.73\ m/s^2$

$a_s$ = acceleration of Stan = $3.9\ m/s^2$

<u>Assumptions:</u>

$v_k$ = final speed of Kathy when see catches Stan
$v_s$ = final speed of Stan when Kathy catches him
$s_k$ = distance traveled by Kathy to catch Stan
$s_s$ = distance traveled by Stan when Kathy catches him
$t_k$ = time taken by Kathy to catch Stan = $t$
$t_s$ = time interval in which Kathy catches Stan = $t+1$

Part (a):

Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

$\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\$

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

$s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5$

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

$v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3$

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

$v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9$

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

7 0

3 years ago