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seraphim [82]
1 year ago
5

if your vehicle begins to slide or skid in the rear, you should? group of answer choices press the gas pedal and speed up. steer

away from the direction of the rear skid. steer in the direction that the rear is sliding to. hit the breaks and hold on.
Physics
1 answer:
MariettaO [177]1 year ago
7 0

if your vehicle begins to slide or skid in the rear, you should steer in the direction that the rear is sliding to.

<h3>What should you do if the rear end of the vehicle starts to slide or skid?</h3>

In the case above, a person need to Look as well as steer in the direction that the person want the front of the vehicle to go.

Note that  in the case above;

Do stay way from  the brake pedal and never apply the brakes.

  • Steer in the the same direction as the rear of the vehicle that is known to be  sliding. This will give room for the vehicle to go straight instead of going sideways.
  • Then make sure to get ready to steer in the opposite direction if the vehicle is one that begins swerving in the other direction.

Therefore, if your vehicle begins to slide or skid in the rear, you should steer in the direction that the rear is sliding to.

Learn more about slide or skid from

brainly.com/question/2148046

#SPJ1

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Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of th
Vilka [71]

Answer:

The magnitude of the static frictional force is 1200 N

Explanation:

given information :

radius, r = 0.380 m

applied-torque, τ1 = 456 N

The car has a constant velocity, thus the acceleration is zero

α = 0

Στ = I α

τ1 - τ2 = I α

τ2 = counter-torque

τ1 - τ2 = 0

τ1 = τ2

r x F_{s} = τ1

F_{s} = the static frictional force (N)

F_{s} = τ1 /r

  = 456 N/0.380 m

  = 1200 N

7 0
3 years ago
a student calculates experimentally the value of density of an iron as 4.4 gcm³. if the actual density of an iron is 7.6 gcm³, c
ser-zykov [4K]
Hola!

Percentage Error is a measurement of the discrepancy between an observed and a true, or accepted value.

[ refer the attachment. ]

According to Question,

% error = \frac{7.4 - 7.6}{7.6} × 100

= 2.631 % = 2.7 % (approximately.)

hope it helps!

5 0
3 years ago
Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of
kogti [31]

Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

7 0
3 years ago
Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
seraphim [82]

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

7 0
3 years ago
Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.
masya89 [10]
The velocity would be the same for all ramps.
5 0
3 years ago
Read 2 more answers
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