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tatiyna
2 years ago
10

How can I protect my computer from being shot by an MBT

Physics
1 answer:
Pie2 years ago
4 0

Answer: While Technology is Evolving so is Hackers that being said "Norton Vpn" would be beneficial for computer protection.

Explanation:

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Which is the best description of inertia? Inertia is
valentinak56 [21]

Answer:

The best description of inertia is an objects velocity

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3 years ago
The horizontal force exerted between the tires of a 500kg car and the ground is 980N. if the car starts from rest, how far will
Maksim231197 [3]

Answer:

Explanation:

d = ½at²

d = ½(F/m)t²

d = ½(980/500)5²

d = 24.5 m

6 0
2 years ago
Tremendous energy is released by nuclear reactions is a measure of the energy required to bond the nucleus together. Where does
Harlamova29_29 [7]

Answer:

Fission is the opposite of fusion and releases energy only when heavy nuclei are split. As noted in Fusion, energy is released if the products of a nuclear reaction have a greater binding energy per nucleon than the parent nuclei.

The amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only 0.1 percent of the original nuclei is converted to energy. Daughter nucleus, energy, and particles such as neutrons are released as a result of the reaction

5 0
3 years ago
Please help me with this
amm1812

the answer might be 2.3 kilos

3 0
4 years ago
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
4 years ago
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