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garri49 [273]
3 years ago
8

You are driving in rainy weather, on a highway with bushes and trees along the sides of the road. If you are driving too fast an

d your vehicle goes off the road, you should
Physics
2 answers:
maxonik [38]3 years ago
7 0

Answer:

Personally I would pull over and wait the rain out in my car, preferably parked under a tree.

GuDViN [60]3 years ago
7 0

Answer:

when you are driving on the road in rainy weather on a highway with bushes and trees on both the sides and you are moving on the road with high speed and your vehicle goes off the road what you should is, first of all, apply the brake and park on the side of the road and wait for the weather to clear or drive on the road slowly so that vehicle can make better contact with the road  and vehicle don't slip again.

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Based on the information in the table shown, which statement describes what a chemist would find if she analyzed a large piece o
Otrada [13]
A) It would likely contain all the isotopes of nickel shown in the chart in the same proportions.

[only one answer here, everything is wrong other than this ]

Hope this helps!
4 0
3 years ago
Read 2 more answers
N april 1974, steve prefontaine completed a 10-km race in a time of 27 min, 43.6 s. suppose \"pre\" was at the 7.15-km mark at a
aliya0001 [1]

solution:speed is 7.99km ,u=\frac{7990}{(25\times60)}=5.33m/s\\
now let the final velocity after 60s from 7.99km is v.\\
v-u =at\\v=5.33+a\times60\\  =60s+5.33\\now distance travelled during this time will be\\
s_{2}=vt\\
v\times103.6\\
=(60a+5.33)\times103.6\\
=551.84+6216am\\
now total distance travelled  is \\  also\\
s_{1}=ut+(\frac{1}{2})at^{2}      =5.33\times60+(0.5)9\times3600\\
=1800a+319.6m\\
Now the remaining time to complete rest race was\\
t=(27\times60+43.6)60-60(25\times60)\\
=103.6s\\Now, distance travelled during this time will be\\
s_{2}=vt\\
v\times103.6\\
=(60a+5.33)\times103.6\\
=551.84+6216am\\
now total distance travelled  is \\
s_{1}+s_{2}+7990=10000\\
1800a+319.6+551.84+6216a=2010\\
8016a=2010-871.44\\
8016a=1138.56\\
a=0.142m/s^2

6 0
3 years ago
If an object is thrown upward with an initial velocity of 128 ​ft/second, then its height after t seconds is given by the follow
IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

h = 128t-32t^2....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

\dfrac{dh}{dt}=0

From equation (I)

\dfrac{dh}{dt}=128-64t

128-64t=0

t=\dfrac{128}{64}

t=2\ sec

Now, for maximum height

Put the value of t in equation (I)

h =128\times2-32\times4

h=128\ ft

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

128t-32t^2=0

t(128-32t)=0

128=32t

t=4\ sec

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0
3 years ago
A concrete column has a diameter of 380 mm and a length of 2.6 m . if the density (mass/volume) of concrete is 2.45 mg/m3, deter
grigory [225]

The volume of the column is

(π) · (r²) · (length) =

(π) · (0.19 meter)² · (2.6 meters) =

(π) · (0.036 m²) · (2.6 m) =

0.294 m³ .

The density is 2,450 kg/m³ (VERY very dense, heavy concrete)

so the weight of the column is (mass)·(gravity) or

(density) · (volume) · (gravity) =

(2,450 kg/m³) · (0.294 m³) · (9.81 m/s²) =

(2,450 · 0.294 · 9.81) (kg · m³· m) / (m³ · s²) =

7,066 kg-m/s² = 7,066 Newtons .

But 9.81 Newtons = 2.20462 pounds on Earth (the weight of 1 kilogram of mass), so we have

(7,066 N) · (2.205 pound/9.81 N) =

(7,066 · 2.205 / 9.81) pounds =

1,588 pounds .

5 0
3 years ago
In a local bar, a customer slides an empty beer mug down the counter for a refill. the height of the counter is 1.15 m. the mug
yarga [219]

<span>A.    </span>Let’s say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the components are v0x and v0y. 
<span>v0y = 0 since the customer slides it horizontally so applied force is in the x component only.

<span>The equations for horizontal and vertical projectile motion are:
x = x0 + v0x t 
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>

Setting the origin to be the end corner of the counter so that x0=0 and y0=0, hence:

x = v0x t

y = - 1/2 g t^2 

Given value are: x=1.50m and y=-1.15m (y is negative since mug is going down)

<span>1.50m = v0x t    ---->  v0x= 1.50/t</span>

<span>-1.15m = -(1/2) (9.81) t^2    -----> t =0.4842 s</span>

Calculating for v0x:

v0x = 3.10 m/s

<span>B.    </span>v0x is constant since there are no other horizontal forces so, v0x=vx=3.10m/s

vy can be calculated from the formula:

<span>vy = v0y + at         where a=-g (negative since going down)</span>

vy = -gt = -9.81 (0.4842)

vy = -4.75 m/s

 

Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy

tan(90-Ø )= 3.1/4.75

Ø = 56.87˚<span> below the horizontal</span>

4 0
4 years ago
Read 2 more answers
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