Answer:
21 m
Explanation:
The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using
![d=vt](https://tex.z-dn.net/?f=d%3Dvt)
where
d is the distance covered
v is the speed
t is the time
The frog in this problem has a speed of
v = 2.1 m/s
and therefore, after t = 10 s, the distance it covered is
![d=(2.1)(10)=21 m](https://tex.z-dn.net/?f=d%3D%282.1%29%2810%29%3D21%20m)
Answer:
between 10 and 15 percent
Explanation:
How to put your load
- First load the heavy
The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.
In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.
- Hold your load
After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.
- Trailer weight
To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.
GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.
Explanation:
The equation of motion of an object is given by :
![h(t)=-16t^2+112t+128](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B112t%2B128)
Where
t is the time in seconds
We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0
So,
![-16t^2+112t+128=0](https://tex.z-dn.net/?f=-16t%5E2%2B112t%2B128%3D0)
![-t^2+7t+8=0](https://tex.z-dn.net/?f=-t%5E2%2B7t%2B8%3D0)
On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.
Answer:
(a)
![x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km](https://tex.z-dn.net/?f=x_1%3D17.68km%5C%5Cy_1%3D-17.68km%5C%5Cx_2%3D20km%5C%5Cy_2%3D34.64km)
(b) ![r=41.32km](https://tex.z-dn.net/?f=r%3D41.32km)
![\alpha =24.23^o](https://tex.z-dn.net/?f=%5Calpha%20%3D24.23%5Eo)
Explanation:
Let us take the north direction to be the positive y-axis and the east to be positive x-axis.
First day:
25.0 km southeast, which implies
south of east. The y-component will be negative and the x-component will be positive.
![x_1=25cos45^o=17.68km](https://tex.z-dn.net/?f=x_1%3D25cos45%5Eo%3D17.68km)
![y_1=-25sin45^o=-17.68km](https://tex.z-dn.net/?f=y_1%3D-25sin45%5Eo%3D-17.68km)
Second day:
She starts off at the stopping point of last day. This time, both the y- and x-components are positive.
![x_2=40cos60^o=20km](https://tex.z-dn.net/?f=x_2%3D40cos60%5Eo%3D20km)
![y_2=40sin60^o=34.64km](https://tex.z-dn.net/?f=y_2%3D40sin60%5Eo%3D34.64km)
Therefore, total displacements:
![x=x_1+x_2=(17.68+20)km=37.68km](https://tex.z-dn.net/?f=x%3Dx_1%2Bx_2%3D%2817.68%2B20%29km%3D37.68km)
![y=y_1+y_2=(-17.68+34.64)km=16.96km](https://tex.z-dn.net/?f=y%3Dy_1%2By_2%3D%28-17.68%2B34.64%29km%3D16.96km)
Magnitude of displacements,
![r=\sqrt{x^2+y^2}=41.32km](https://tex.z-dn.net/?f=r%3D%5Csqrt%7Bx%5E2%2By%5E2%7D%3D41.32km)
Direction,
![\alpha =tan^{-1}(\frac{y}{x})=24.23^o](https://tex.z-dn.net/?f=%5Calpha%20%3Dtan%5E%7B-1%7D%28%5Cfrac%7By%7D%7Bx%7D%29%3D24.23%5Eo)
True,
Explanation: because it’s the same reason that lighting touches the ground/ goes up to the sky, because it is trying to balance out its charges